MCQs

Question 1

A test charge of
+2μC experiences a force of
4×10 ^−3N at a certain point in an electric field. If the test charge is replaced with
+4μC, what will be the new electric field intensity?

A) 2000N/C
B) 4000N/C
C) 1000N/C
D) 500N/C

Answer 1

e = F/q
where:

E is the electric field intensity,
F is the force experienced by the charge,
q is the charge.
Step 1: Calculate the initial electric field
Given:

F= 4 × 10^−3N,
q=2 μC = 2×10 ^−6C.

E= 4×10^−3/2 ×10^−6
E=2000N/C
Step 2: Determine the new electric field intensity
Since the electric field intensity (E) at a point is independent of the charge placed there, it remains the same regardless of the test charge used.

Thus, the new electric field intensity remains:

2000N/C

Question 2

Two point charges,
+3μC and −3μC, are placed 6 cm apart. What is the net electric field at the midpoint of the two charges?

A) 30×10 ^5N/C towards the negative charge
B) 15×10 ^5N/C away from both charges
C) 10×10 ^ 5 N/C towards the positive charge
D) 20×10 ^5 N/C towards the negative charge

Answer 2

E

Solution: The electric field due to a positive charge points away, while for a negative charge, it points towards the charge. The magnitudes of the electric fields due to each charge at the midpoint are equal and add up in the same direction (towards the negative charge).

Using the formula:

E = kQ/r^2
where
k = 9 × 10^9
Q=3μC=3×10 ^−6C, and
r=3×10 ^−2m, we find:

(
9
×
10
9
)
(
3
×
10
−
6
)
(
3
×
10
−
2
)
2
=
30
×
10
5
 
N
/
C
E=
(3×10
−2
)
2

(9×10
9
)(3×10
−6
)
​
=30×10
5
N/C
Correct Answer: A)
30×10 ^5N/C towards the negative charge

Question 3

A charge of
+
5
μ
C
+5μC is placed at the origin. A second charge of
−
5
μ
C
−5μC is introduced at
x
=
10
x=10 cm. What happens to the electric field at
x
=
5
x=5 cm?

A) It increases towards the negative charge
B) It decreases due to cancellation
C) It remains the same
D) It reverses direction

Answer 3

Solution: The first charge alone creates an electric field. However, when the second charge is introduced, their fields interact. Since they are opposite charges, the field due to each charge at
x
=
5
x=5 cm cancels out partially, leading to a weaker net field.

Correct Answer: B) It decreases due to cancellatio

Question 4

A test charge of
+
3
μ
C
+3μC is placed in an electric field and experiences a force of
6
×
10
−
3
N
6×10
−3
N. If the test charge is replaced with
+
6
μ
C
+6μC, what will be the new force experienced by it?

A)
6
×
10
−
3
N
6×10
−3
N
B)
12
×
10
−
3
N
12×10
−3
N
C)
3
×
10
−
3
N
3×10
−3
N
D)
9
×
10
−
3
N
9×10
−3
N

Answer 4

E

Solution

Electric field is given by:

F
q
E=
q
F
​

For the first charge:

6
×
10
−
3
N
3
×
10
−
6
C
=
2000
 
N
/
C
E=
3×10
−6
C
6×10
−3
N
​
=2000N/C
Since the electric field is independent of the test charge, when we place a new charge
q
′
=
6
μ
C
q
′
=6μC, the force is given by:

F
′
=
E
q
′
=
(
2000
)
×
(
6
×
10
−
6
)
F
′
=Eq
′
=(2000)×(6×10
−6
)
F
′
=
12
×
10
−
3
N
F
′
=12×10
−3
N
Correct Answer: B)
12
×
10
−
3
N
12×10
−3
N

Question 5

An electric field of
5000
N
/
C
5000N/C is created by a charge. If a
+
2
μ
C
+2μC charge is placed in the field, what force does it experience?

A)
10
×
10
−
3
N
10×10
−3
N
B)
5
×
10
−
3
N
5×10
−3
N
C)
15
×
10
−
3
N
15×10
−3
N
D)
1
×
10
−
3
N
1×10
−3
N

Answer 5

F

Solution

Using:

E
q
F=Eq
F
=
(
5000
)
×
(
2
×
10
−
6
)
F=(5000)×(2×10
−6
)
F
=
10
×
10
−
3
N
F=10×10
−3
N
Correct Answer: A)
10
×
10
−
3
N
10×10
−3
N

Question 6

Two point charges
+
4
μ
C
+4μC and
−
4
μ
C
−4μC are placed 8 cm apart. What is the magnitude of the net electric field at the midpoint between them?

A)
45
×
10
4
N
/
C
45×10
4
N/C
B)
90
×
10
4
N
/
C
90×10
4
N/C
C)
30
×
10
4
N
/
C
30×10
4
N/C
D)
60
×
10
4
N
/
C
60×10
4
N/C

Answer 6

E

Electric field due to a point charge is:

k
Q
r
2
E=
r
2

kQ
​

where
k
=
9
×
10
9
k=9×10
9
,
Q
=
4
μ
C
=
4
×
10
−
6
C
Q=4μC=4×10
−6
C, and
r
=
4
×
10
−
2
m
r=4×10
−2
m (since the midpoint is at 4 cm from each charge).

(
9
×
10
9
)
(
4
×
10
−
6
)
(
4
×
10
−
2
)
2
E=
(4×10
−2
)
2

(9×10
9
)(4×10
−6
)
​

(
36
×
10
3
)
16
×
10
−
4
E=
16×10
−4

(36×10
3
)
​

22.5
×
10
4
N
/
C
E=22.5×10
4
N/C
Since both fields add up in the same direction, the net field is:

E
net
=
2
E
=
45
×
10
4
N
/
C
E
net
​
=2E=45×10
4
N/C
Correct Answer: A)
45
×
10
4
N
/
C
45×10
4
N/C

Question 7

A positive test charge experiences a force of
8
×
10
−
3
N
8×10
−3
N in the positive x-direction due to a charge. If another identical test charge is placed at the same point, in what direction will it experience the force?

A) Negative x-direction
B) Positive x-direction
C) Negative y-direction
D) Positive y-direction

Answer 7

Since electric field is a vector, the direction of force on a positive test charge is always along the field lines. If the first test charge experienced a force in the positive x-direction, any identical test charge at that position will also experience a force in the same direction.

Correct Answer: B) Positive x-direction

Question 8

A charge of
+
5
μ
C
+5μC is placed at the origin. A second charge of
−
5
μ
C
−5μC is introduced at
x
=
10
x=10 cm. What happens to the electric field at
x
=
5
x=5 cm?

A) It increases towards the negative charge
B) It decreases due to cancellation
C) It remains the same
D) It reverses direction

Answer 8

Initially, the field at
x
=
5
x=5 cm was only due to the
+
5
μ
C
+5μC charge. When the
−
5
μ
C
−5μC charge is introduced, it creates an opposite field direction. Since both charges are equal and opposite, the fields at
x
=
5
x=5 cm partially cancel out, leading to a decrease in total electric field.

Correct Answer: B) It decreases due to cancellation

Question 9

A
+
3
μ
C
+3μC charge is placed at the origin. Another
+
6
μ
C
+6μC charge is introduced at
x
=
10
x=10 cm. How does the field at
x
=
5
x=5 cm change?

A) It increases in the positive x-direction
B) It decreases in the positive x-direction
C) It reverses direction
D) It remains the same

Answer 9

Solution

Before the
+
6
μ
C
+6μC charge was introduced, the electric field at
x
=
5
x=5 cm was due to the
+
3
μ
C
+3μC charge, pointing away from it. After the
+
6
μ
C
+6μC charge is introduced, it adds its own field, also pointing away from it, in the same direction.

Since the additional charge is larger, the total field increases.

Correct Answer: A) It increases in the positive x-direction

Question 10

Two charges,
+
3
μ
C
+3μC and
−
2
μ
C
−2μC, are placed 8 cm apart. What is the net electric field at a point midway between them?

A)
15
×
10
4
N
/
C
15×10
4
N/C towards the negative charge
B)
10
×
10
4
N
/
C
10×10
4
N/C away from both charges
C)
5
×
10
4
N
/
C
5×10
4
N/C towards the positive charge
D)
20
×
10
4
N
/
C
20×10
4
N/C towards the negative charge

Answer 10

E

Using the formula for the electric field:

k
Q
r
2
E=
r
2

kQ
​

where:

9
×
10
9
k=9×10
9

3
μ
C
Q=3μC and
−
2
μ
C
−2μC
r
=
4
c
m
=
0.04
m
r=4cm=0.04m (since midpoint is 4 cm from each charge)
E
1
=
(
9
×
10
9
)
(
3
×
10
−
6
)
(
0.04
)
2
E
1
​
=
(0.04)
2

(9×10
9
)(3×10
−6
)
​

E
1
=
27
×
10
3
1.6
×
10
−
3
=
16.875
×
10
4
N
/
C
E
1
​
=
1.6×10
−3

27×10
3

​
=16.875×10
4
N/C
E
2
=
(
9
×
10
9
)
(
2
×
10
−
6
)
(
0.04
)
2
E
2
​
=
(0.04)
2

(9×10
9
)(2×10
−6
)
​

E
2
=
18
×
10
3
1.6
×
10
−
3
=
11.25
×
10
4
N
/
C
E
2
​
=
1.6×10
−3

18×10
3

​
=11.25×10
4
N/C
Since
+
3
μ
C
+3μC pushes the test charge away and
−
2
μ
C
−2μC pulls it towards, the net field is:

E
net
=
16.875
−
11.25
=
5.625
×
10
4
N
/
C
E
net
​
=16.875−11.25=5.625×10
4
N/C
Direction: Towards the negative charge.

Correct Answer: A)
15
×
10
4
N
/
C
15×10
4
N/C towards the negative charge

Question 11

Three charges
+
4
μ
C
+4μC,
−
2
μ
C
−2μC, and
+
5
μ
C
+5μC are placed at three different points in a straight line. What is the net electric field at a point between the first two charges?

A) It depends on the distance and signs of the charges
B) The field will always be zero
C) The field will always point towards the largest charge
D) The field will only depend on the closest charge

Answer 11

By the superposition principle, the net electric field is a sum of individual fields. Since electric fields follow vector addition:

If the test point is closer to a strong charge, its effect dominates.
The net direction depends on the charge signs.
Thus, the answer is:

Correct Answer: A) It depends on the distance and signs of the charges

Question 12

Two charges,
+
4
μ
C
+4μC and
+
9
μ
C
+9μC, are placed 10 cm apart. Find the point along the line joining them where the electric field is zero.

Answer 12

E

The electric field cancels out when:

E
1
=
E
2
E
1
​
=E
2
​

Using the formula for the electric field:

k
Q
r
2
E=
r
2

kQ
​

Let the point where
E
=
0
E=0 be at a distance
x
x from the smaller charge (
+
4
μ
C
+4μC). The distance from the larger charge is
(
10
−
x
)
(10−x) cm.

Setting the electric fields equal:

k
(
4
×
10
−
6
)
x
2
=
k
(
9
×
10
−
6
)
(
10
−
x
)
2
x
2

k(4×10
−6
)
​
=
(10−x)
2

k(9×10
−6
)
​

Canceling
k
k and solving for
x
x:

4
x
2
=
9
(
10
−
x
)
2
x
2

4
​
=
(10−x)
2

9
​

Cross multiplying:

4
(
10
−
x
)
2
=
9
x
2
4(10−x)
2
=9x
2

Expanding:

4
(
100
−
20
x
+
x
2
)
=
9
x
2
4(100−20x+x
2
)=9x
2

400
−
80
x
+
4
x
2
=
9
x
2
400−80x+4x
2
=9x
2

Rearrange:

400
−
80
x
+
4
x
2
−
9
x
2
=
0
400−80x+4x
2
−9x
2
=0
400
−
80
x
−
5
x
2
=
0
400−80x−5x
2
=0
Solving for
x
x using the quadratic formula:

−
(
−
80
)
±
(
−
80
)
2
−
4
(
−
5
)
(
400
)
2
(
−
5
)
x=
2(−5)
−(−80)±
(−80)
2
−4(−5)(400)
​

​

80
±
6400
+
8000
−
10
x=
−10
80±
6400+8000
​

​

80
±
14400
−
10
x=
−10
80±
14400
​

​

80
±
120
−
10
x=
−10
80±120
​

Solving for
x
x:

80
+
120
−
10
=
200
−
10
=
−
20
x=
−10
80+120
​
=
−10
200
​
=−20 (not possible)
x
=
80
−
120
−
10
=
−
40
−
10
=
4
x=
−10
80−120
​
=
−10
−40
​
=4 cm
Thus, the electric field is zero 4 cm away from the
+
4
μ
C
+4μC charge (closer to the smaller charge).

Question 13

Two charges,
+
6
μ
C
+6μC and
−
2
μ
C
−2μC, are placed 12 cm apart. Find the point where the electric field is zero.

Answer 13

x

For opposite charges, the zero-field point is always outside the charges (not between them). Let the point be at a distance
x
x from the smaller charge (
−
2
μ
C
−2μC).

Using:

k
(
6
×
10
−
6
)
(
x
+
12
)
2
=
k
(
2
×
10
−
6
)
x
2
(x+12)
2

k(6×10
−6
)
​
=
x
2

k(2×10
−6
)
​

Cancel
k
k:

6
(
x
+
12
)
2
=
2
x
2
(x+12)
2

6
​
=
x
2

2
​

Cross multiply:

6
x
2
=
2
(
x
+
12
)
2
6x
2
=2(x+12)
2

Expanding:

6
x
2
=
2
(
x
2
+
24
x
+
144
)
6x
2
=2(x
2
+24x+144)
6
x
2
=
2
x
2
+
48
x
+
288
6x
2
=2x
2
+48x+288
Rearrange:

6
x
2
−
2
x
2
−
48
x
−
288
=
0
6x
2
−2x
2
−48x−288=0
4
x
2
−
48
x
−
288
=
0
4x
2
−48x−288=0
Divide by 4:

x
2
−
12
x
−
72
=
0
x
2
−12x−72=0
Solving using the quadratic formula:

−
(
−
12
)
±
(
−
12
)
2
−
4
(
1
)
(
−
72
)
2
(
1
)
x=
2(1)
−(−12)±
(−12)
2
−4(1)(−72)
​

​

12
±
144
+
288
2
x=
2
12±
144+288
​

​

12
±
432
2
x=
2
12±
432
​

​

Approximating
432
≈
20.78
432
​
≈20.78:

12
±
20.78
2
x=
2
12±20.78
​

Solving for
x
x:

12
+
20.78
2
=
32.78
2
=
16.39
x=
2
12+20.78
​
=
2
32.78
​
=16.39 cm
x
=
12
−
20.78
2
=
−
8.78
2
=
−
4.39
x=
2
12−20.78
​
=
2
−8.78
​
=−4.39 (not possible)
Thus, the electric field is zero 16.39 cm to the left of the
−
2
μ
C
−2μC charge.

Key Takeaways
For like charges, the zero-field point is between them, closer to the smaller charge.
For opposite charges, the zero-field point is outside, farther from the smaller charge.
Quadratic equations are often used to solve for the location.
Would you like more advanced cases (e.g., three charges)? 🚀

Question 14

Three point charges
+
3
μ
C
+3μC,
+
6
μ
C
+6μC, and
+
9
μ
C
+9μC are placed along a straight line at equal distances of 10 cm apart. Find the point where the electric field is zero.

Answer 14

d

Solution

We analyze each charge and its contribution to the total electric field.

Let the charges be at:

Q
1
=
+
3
μ
C
Q
1
​
=+3μC at
x
=
0
x=0,
Q
2
=
+
6
μ
C
Q
2
​
=+6μC at
x
=
10
x=10 cm,
Q
3
=
+
9
μ
C
Q
3
​
=+9μC at
x
=
20
x=20 cm.
Since all charges are positive, the zero-field point must be between the two smaller charges (
Q
1
Q
1
​
and
Q
2
Q
2
​
) or between
Q
2
Q
2
​
and
Q
3
Q
3
​
.

Step 1: Assume the Zero-Field Point is Between
Q
1
Q
1
​
and
Q
2
Q
2
​

Let the point be at
x
=
d
x=d cm from
Q
1
Q
1
​
(
d
<
10
d<10).

At this point:

E
1
=
E
2
E
1
​
=E
2
​

k
(
3
×
10
−
6
)
d
2
=
k
(
6
×
10
−
6
)
(
10
−
d
)
2
d
2

k(3×10
−6
)
​
=
(10−d)
2

k(6×10
−6
)
​

Cancel
k
k:

3
d
2
=
6
(
10
−
d
)
2
d
2

3
​
=
(10−d)
2

6
​

Cross multiply:

3
(
10
−
d
)
2
=
6
d
2
3(10−d)
2
=6d
2

Expanding:

3
(
100
−
20
d
+
d
2
)
=
6
d
2
3(100−20d+d
2
)=6d
2

300
−
60
d
+
3
d
2
=
6
d
2
300−60d+3d
2
=6d
2

Rearrange:

300
−
60
d
−
3
d
2
=
0
300−60d−3d
2
=0
3
d
2
+
60
d
−
300
=
0
3d
2
+60d−300=0
Dividing by 3:

d
2
+
20
d
−
100
=
0
d
2
+20d−100=0
Solving using the quadratic formula:

−
20
±
(
20
)
2
−
4
(
1
)
(
−
100
)
2
(
1
)
d=
2(1)
−20±
(20)
2
−4(1)(−100)
​

​

−
20
±
400
+
400
2
d=
2
−20±
400+400
​

​

−
20
±
800
2
d=
2
−20±
800
​

​

Approximating
800
≈
28.28
800
​
≈28.28:

−
20
±
28.28
2
d=
2
−20±28.28
​

Solving for
d
d:

−
20
+
28.28
2
=
8.28
2
=
4.14
d=
2
−20+28.28
​
=
2
8.28
​
=4.14 cm
d
=
−
20
−
28.28
2
=
−
48.28
2
=
−
24.14
d=
2
−20−28.28
​
=
2
−48.28
​
=−24.14 cm (not possible)
Thus, the electric field is zero at 4.14 cm from
Q
1
Q
1
​
, between
Q
1
Q
1
​
and
Q
2
Q
2
​
.

Question 15

Three point charges
+
5
μ
C
+5μC,
−
3
μ
C
−3μC, and
+
2
μ
C
+2μC are placed at
x
=
0
x=0,
x
=
8
x=8 cm, and
x
=
16
x=16 cm. Find the point where the electric field is zero.

Answer 15

d

Solution

Since we have opposite charges, the zero-field point is likely outside the system or between two charges.
The negative charge will attract a test charge, while positive charges push it away.
Step 1: Assume the Zero-Field Point is Between
+
5
μ
C
+5μC and
−
3
μ
C
−3μC

Let it be at
x
=
d
x=d cm from
+
5
μ
C
+5μC (before
−
3
μ
C
−3μC, meaning
d
<
8
d<8).

Using:

k
(
5
×
10
−
6
)
d
2
=
k
(
3
×
10
−
6
)
(
8
−
d
)
2
d
2

k(5×10
−6
)
​
=
(8−d)
2

k(3×10
−6
)
​

Cancel
k
k:

5
d
2
=
3
(
8
−
d
)
2
d
2

5
​
=
(8−d)
2

3
​

Cross multiply:

5
(
8
−
d
)
2
=
3
d
2
5(8−d)
2
=3d
2

Expanding:

5
(
64
−
16
d
+
d
2
)
=
3
d
2
5(64−16d+d
2
)=3d
2

320
−
80
d
+
5
d
2
=
3
d
2
320−80d+5d
2
=3d
2

Rearrange:

320
−
80
d
+
5
d
2
−
3
d
2
=
0
320−80d+5d
2
−3d
2
=0
2
d
2
−
80
d
+
320
=
0
2d
2
−80d+320=0
Dividing by 2:

d
2
−
40
d
+
160
=
0
d
2
−40d+160=0
Solving using the quadratic formula:

−
(
−
40
)
±
(
−
40
)
2
−
4
(
1
)
(
160
)
2
(
1
)
d=
2(1)
−(−40)±
(−40)
2
−4(1)(160)
​

​

40
±
1600
−
640
2
d=
2
40±
1600−640
​

​

40
±
960
2
d=
2
40±
960
​

​

Approximating
960
≈
31
960
​
≈31:

40
±
31
2
d=
2
40±31
​

Solving for
d
d:

40
+
31
2
=
71
2
=
35.5
d=
2
40+31
​
=
2
71
​
=35.5 cm (not valid, as it must be within 8 cm)
d
=
40
−
31
2
=
9
2
=
4.5
d=
2
40−31
​
=
2
9
​
=4.5 cm
Thus, the electric field is zero at
x
=
4.5
x=4.5 cm, between
+
5
μ
C
+5μC and
−
3
μ
C
−3μC.

Key Takeaways
For like charges, the zero-field point is between two smaller charges.
For opposite charges, the zero-field point is usually outside or between them.
Superposition is crucial—sum fields vectorially.
Quadratic equations help solve most problems.
Would you like a more advanced configuration, like an equilateral triangle or 3D case? 🚀

Question 16

A
+
5
μ
C
+5μC charge creates an electric field. At a distance of 4 cm, the field strength is
2.8
×
10
6
N
/
C
2.8×10
6
N/C. What will be the field strength at 2 cm?

A)
5.6
×
10
6
N
/
C
5.6×10
6
N/C
B)
11.2
×
10
6
N
/
C
11.2×10
6
N/C
C)
1.4
×
10
6
N
/
C
1.4×10
6
N/C
D)
0.7
×
10
6
N
/
C
0.7×10
6
N/C

Answer 16

Since
E
∝
1
r
2
E∝
r
2

1
​
, we use:

E
2
=
E
1
×
(
r
1
r
2
)
2
E
2
​
=E
1
​
×(
r
2
​

r
1
​

​
)
2

E
2
=
(
2.8
×
10
6
)
×
(
4
2
)
2
E
2
​
=(2.8×10
6
)×(
2
4
​
)
2

E
2
=
(
2.8
×
10
6
)
×
(
2
2
)
E
2
​
=(2.8×10
6
)×(2
2
)
E
2
=
(
2.8
×
10
6
)
×
4
E
2
​
=(2.8×10
6
)×4
E
2
=
11.2
×
10
6
N
/
C
E
2
​
=11.2×10
6
N/C
Correct Answer: B)
11.2
×
10
6
N
/
C
11.2×10
6
N/C

Question 17

If the electric field at a distance
r
r from a charge is
6
×
10
5
N
/
C
6×10
5
N/C, at what distance will the field be
1.5
×
10
5
N
/
C
1.5×10
5
N/C?

A)
2
r
2r
B)
r
2
2
r
​

C)
r
4
4
r
​

D)
4
r
4r

Answer 17

Solution

Using the inverse square law:

E
2
=
E
1
×
(
r
1
r
2
)
2
E
2
​
=E
1
​
×(
r
2
​

r
1
​

​
)
2

1.5
×
10
5
=
6
×
10
5
×
(
r
r
2
)
2
1.5×10
5
=6×10
5
×(
r
2
​

r
​
)
2

Dividing both sides by
6
×
10
5
6×10
5
:

1.5
6
=
(
r
r
2
)
2
6
1.5
​
=(
r
2
​

r
​
)
2

1
4
=
(
r
r
2
)
2
4
1
​
=(
r
2
​

r
​
)
2

Taking square root:

1
2
=
r
r
2
2
1
​
=
r
2
​

r
​

r
2
=
2
r
r
2
​
=2r
Correct Answer: A)
2
r
2r

Question 18

Two charges,
+
5
μ
C
+5μC and
−
5
μ
C
−5μC, are separated by 8 cm. What is the dipole moment?

A)
4.0
×
10
−
7
C
⋅
m
4.0×10
−7
C⋅m
B)
2.0
×
10
−
7
C
⋅
m
2.0×10
−7
C⋅m
C)
1.6
×
10
−
7
C
⋅
m
1.6×10
−7
C⋅m
D)
5.0
×
10
−
7
C
⋅
m
5.0×10
−7
C⋅m

Answer 18

p

Solution: Using:

q
d
p=qd
p
=
(
5
×
10
−
6
C
)
×
(
8
×
10
−
2
m
)
p=(5×10
−6
C)×(8×10
−2
m)
p
=
4.0
×
10
−
7
C
⋅
m
p=4.0×10
−7
C⋅m
Correct Answer: A)
4.0
×
10
−
7
C
⋅
m
4.0×10
−7
C⋅m

Question 19

orque on a Dipole

A dipole with a moment of
3
×
10
−
9
C
⋅
m
3×10
−9
C⋅m is placed in a uniform electric field of
5
×
10
5
N
/
C
5×10
5
N/C at an angle of
30
∘
30
∘
. What is the torque on the dipole?

A)
6.0
×
10
−
4
N
⋅
m
6.0×10
−4
N⋅m
B)
7.5
×
10
−
4
N
⋅
m
7.5×10
−4
N⋅m
C)
3.75
×
10
−
4
N
⋅
m
3.75×10
−4
N⋅m
D)
1.5
×
10
−
4
N
⋅
m
1.5×10
−4
N⋅m

Answer 19

τ

Solution: Using:

p
E
sin
⁡
θ
τ=pEsinθ
τ
=
(
3
×
10
−
9
)
×
(
5
×
10
5
)
×
sin
⁡
30
∘
τ=(3×10
−9
)×(5×10
5
)×sin30
∘

Since
sin
⁡
30
∘
=
0.5
sin30
∘
=0.5:

(
3
×
10
−
9
)
×
(
5
×
10
5
)
×
0.5
τ=(3×10
−9
)×(5×10
5
)×0.5
τ
=
(
3
×
5
×
0.5
)
×
10
−
9
+
5
τ=(3×5×0.5)×10
−9+5

(
7.5
)
×
10
−
4
τ=(7.5)×10
−4

Correct Answer: B)
7.5
×
10
−
4
N
⋅
m
7.5×10
−4
N⋅m

Question 20

A dipole has a dipole moment of
2
×
10
−
9
C
⋅
m
2×10
−9
C⋅m. Find the electric field at a distance of 5 cm from the center on the equatorial line.

A)
1.44
×
10
4
N
/
C
1.44×10
4
N/C
B)
2.88
×
10
4
N
/
C
2.88×10
4
N/C
C)
3.60
×
10
4
N
/
C
3.60×10
4
N/C
D)
4.32
×
10
4
N
/
C
4.32×10
4
N/C

Answer 20

E

Solution: Using:

E
equatorial
=
1
4
π
ε
0
⋅
p
r
3
E
equatorial
​
=
4πε
0
​

1
​
⋅
r
3

p
​

9
×
10
9
×
2
×
10
−
9
(
0.05
)
3
E=
(0.05)
3

9×10
9
×2×10
−9

​

18
×
10
0
1.25
×
10
−
4
E=
1.25×10
−4

18×10
0

​

1.44
×
10
4
N
/
C
E=1.44×10
4
N/C
Correct Answer: A)
1.44
×
10
4
N
/
C
1.44×10
4
N/C

Question 21

Dipole Moment of a Water (H₂O) Molecule

Water has polar covalent bonds, and the oxygen-hydrogen bond has a dipole due to the difference in electronegativity.

Given:

Partial charge on oxygen:
q
=
1.85
×
10
−
19
q=1.85×10
−19
C
Bond length between O and H:
d
=
0.958
×
10
−
10
d=0.958×10
−10
m
Water has two bonds with an angle of 104.5°, so we use vector addition.
Each O-H bond contributes to the dipole moment. The net dipole moment of water is:

Answer 21

p
net
​
=1.85×10
−19
×0.958×10
−10
=1.77×10
−29
C\cdotpm
After considering the vector sum, the total dipole moment of water is about 1.85 D.

Question 22

q

Dipole Moment of Hydrogen Chloride (HCl)

HCl has a polar bond because chlorine is much more electronegative than hydrogen.

Given:

3.336
×
10
−
19
q=3.336×10
−19
C
d
=
1.27
×
10
−
10
d=1.27×10
−10
m

Answer 22

p

p=(3.336×10
−19
)×(1.27×10
−10
)
p
=
4.24
×
10
−
30
C
\cdotp
m
p=4.24×10
−30
C\cdotpm
Converting to Debye:

4.24
×
10
−
30
3.336
×
10
−
30
=
1.27
D
p=
3.336×10
−30

4.24×10
−30

​
=1.27D
So, HCl has a dipole moment of 1.27 D, making it polar.

Question 23

A dipole with a moment
p
=
4
×
10
−
9
C
⋅
m
p=4×10
−9
C⋅m is placed in space. At a distance
r
r, the electric field is
2.5
×
10
4
N
/
C
2.5×10
4
N/C. What will the field be at twice the distance
(
2
r
)
(2r)?

A)
1.25
×
10
4
N
/
C
1.25×10
4
N/C
B)
6.25
×
10
3
N
/
C
6.25×10
3
N/C
C)
3.125
×
10
3
N
/
C
3.125×10
3
N/C
D)
5.0
×
10
3
N
/
C
5.0×10
3
N/C

Answer 23

Question 24

A dipole with
p
=
4
×
10
−
9
C
⋅
m
p=4×10
−9
C⋅m is placed in a uniform
5
×
10
5
N
/
C
5×10
5
N/C field at an angle of
90
∘
90
∘
. What is the torque?

A)
2.0
×
10
−
3
N
⋅
m
2.0×10
−3
N⋅m
B)
2.0
×
10
−
4
N
⋅
m
2.0×10
−4
N⋅m
C)
1.0
×
10
−
3
N
⋅
m
1.0×10
−3
N⋅m
D)
4.0
×
10
−
3
N
⋅
m
4.0×10
−3
N⋅m

Answer 24

τ

Solution Using:

p
E
sin
⁡
θ
τ=pEsinθ
τ
=
(
4
×
10
−
9
)
×
(
5
×
10
5
)
×
sin
⁡
90
∘
τ=(4×10
−9
)×(5×10
5
)×sin90
∘

Since
sin
⁡
90
∘
=
1
sin90
∘
=1,

2.0
×
10
−
3
N
⋅
m
τ=2.0×10
−3
N⋅m
Correct Answer: A)
2.0
×
10
−
3
N
⋅
m
2.0×10
−3
N⋅m

Question 25

A dipole with p = 3 × 10 − 9 C ⋅ m p=3×10 −9 C⋅m is aligned at 60 ∘ 60 ∘ to an E = 4 × 10 5 N / C E=4×10 5 N/C field. What is its potential energy? A) − 6.0 × 10 − 4 J −6.0×10 −4 J B) − 4.5 × 10 − 4 J −4.5×10 −4 J C) − 3.0 × 10 − 4 J −3.0×10 −4 J D) − 2.0 × 10 − 4 J −2.0×10 −4 J

Answer 25

Solution Using: U = − p E cos ⁡ θ U=−pEcosθ U = − ( 3 × 10 − 9 ) × ( 4 × 10 5 ) × cos ⁡ 60 ∘ U=−(3×10 −9 )×(4×10 5 )×cos60 ∘ Since cos ⁡ 60 ∘ = 0.5 cos60 ∘ =0.5, U = − ( 3 × 10 − 9 × 4 × 10 5 × 0.5 ) U=−(3×10 −9 ×4×10 5 ×0.5) U = − 6.0 × 10 − 4 J U=−6.0×10 −4 J Correct Answer: A) − 6.0 × 10 − 4 J −6.0×10 −4 J

Question 26

A dipole with a moment of 2.5 × 10 − 9 C ⋅ m 2.5×10 −9 C⋅m is placed in a uniform electric field of 3 × 10 5 N / C 3×10 5 N/C at an angle of 45 ∘ 45 ∘ . (a) What is the torque acting on the dipole? (b) At what angle will the torque be maximum?

Answer 26

Solution We use the torque formula: τ = p E sin ⁡ θ τ=pEsinθ τ = ( 2.5 × 10 − 9 ) × ( 3 × 10 5 ) × sin ⁡ 45 ∘ τ=(2.5×10 −9 )×(3×10 5 )×sin45 ∘ Since sin ⁡ 45 ∘ = 2 2 ≈ 0.707 sin45 ∘ = 2 2 ​ ​ ≈0.707, τ = ( 2.5 × 10 − 9 ) × ( 3 × 10 5 ) × ( 0.707 ) τ=(2.5×10 −9 )×(3×10 5 )×(0.707) τ = 5.3 × 10 − 4 N ⋅ m τ=5.3×10 −4 N⋅m (b) Maximum Torque Condition τ max ⁡ = p E sin ⁡ 90 ∘ = p E τ max ​ =pEsin90 ∘ =pE So, the torque is maximum at 90 ∘ 90 ∘ .

Question 27

A dipole with p = 1.8 × 10 − 9 C ⋅ m p=1.8×10 −9 C⋅m is placed in an electric field E = 4 × 10 5 N / C E=4×10 5 N/C. (a) What is the potential energy when the dipole is aligned at 120 ∘ 120 ∘ ? (b) What is the change in energy if the dipole rotates to 30 ∘ 30 ∘ ?

Answer 27

Solution The potential energy is given by: U = − p E cos ⁡ θ U=−pEcosθ (a) At θ = 120 ∘ θ=120 ∘ Since cos ⁡ 120 ∘ = − 0.5 cos120 ∘ =−0.5, U = − ( 1.8 × 10 − 9 ) × ( 4 × 10 5 ) × ( − 0.5 ) U=−(1.8×10 −9 )×(4×10 5 )×(−0.5) U = 3.6 × 10 − 4 J U=3.6×10 −4 J (b) At θ = 30 ∘ θ=30 ∘ Since cos ⁡ 30 ∘ = 0.866 cos30 ∘ =0.866, U ′ = − ( 1.8 × 10 − 9 ) × ( 4 × 10 5 ) × ( 0.866 ) U ′ =−(1.8×10 −9 )×(4×10 5 )×(0.866) U ′ = − 6.2 × 10 − 4 J U ′ =−6.2×10 −4 J Change in Energy Δ U = U ′ − U ΔU=U ′ −U Δ U = ( − 6.2 × 10 − 4 ) − ( 3.6 × 10 − 4 ) ΔU=(−6.2×10 −4 )−(3.6×10 −4 ) Δ U = − 9.8 × 10 − 4 J ΔU=−9.8×10 −4 J Interpretation: The dipole has lost energy as it moved to a more stable position.

Question 28

A water molecule, with a dipole moment of 6.1 × 10 − 30 C ⋅ m 6.1×10 −30 C⋅m, is subjected to an external field of 2.5 × 10 6 N / C 2.5×10 6 N/C. (a) What is the maximum torque the molecule can experience? (b) How much work is required to rotate it from 0 ∘ 0 ∘ to 180 ∘ 180 ∘ ?

Answer 28

Solution (a) Maximum Torque Since torque is maximum at θ = 90 ∘ θ=90 ∘ , τ max ⁡ = p E τ max ​ =pE τ max ⁡ = ( 6.1 × 10 − 30 ) × ( 2.5 × 10 6 ) τ max ​ =(6.1×10 −30 )×(2.5×10 6 ) τ max ⁡ = 1.525 × 10 − 23 N ⋅ m τ max ​ =1.525×10 −23 N⋅m (b) Work Required to Rotate from 0 ∘ 0 ∘ to 180 ∘ 180 ∘ The work done in rotating a dipole is: W = U final − U initial W=U final ​ −U initial ​ At 0 ∘ 0 ∘ , U initial = − p E cos ⁡ 0 ∘ = − p E U initial ​ =−pEcos0 ∘ =−pE. At 180 ∘ 180 ∘ , U final = − p E cos ⁡ 180 ∘ = p E U final ​ =−pEcos180 ∘ =pE. W = p E − ( − p E ) W=pE−(−pE) W = 2 p E W=2pE W = 2 × ( 6.1 × 10 − 30 ) × ( 2.5 × 10 6 ) W=2×(6.1×10 −30 )×(2.5×10 6 ) W = 3.05 × 10 − 23 J W=3.05×10 −23 J

Question 29

A carbon dioxide molecule (CO₂) has no dipole, while a water molecule (H₂O) has a permanent dipole moment. If both molecules are placed in a uniform electric field: (a) Which one will rotate? (b) Which one will experience a net force?

Answer 29

Solution (a) Water ( H 2 O H 2 ​ O) rotates because it has a permanent dipole. (b) Neither CO₂ nor H₂O experience a net force in a uniform electric field. The forces on the positive and negative charges cancel each other out. However, if the field is non-uniform, H₂O will experience both rotation and net force. Key Takeaways Dipoles rotate in an electric field due to torque τ = p E sin ⁡ θ τ=pEsinθ. Dipoles align with the field to minimize potential energy. Work is required to rotate a dipole, calculated as W = 2 p E W=2pE. Uniform fields cause only rotation, but non-uniform fields also exert a force on dipoles. Real-life examples include microwave heating, LCD displays, and molecular physics.

Question 29

A 2 m 2 2m 2 rectangular surface is placed in a uniform electric field of 500 N / C 500N/C. The field makes an angle of 60 ∘ 60 ∘ with the normal to the surface.

Answer 29

Solution Using: Φ E = E A cos ⁡ θ Φ E ​ =EAcosθ Φ E = ( 500 ) × ( 2 ) × cos ⁡ 60 ∘ Φ E ​ =(500)×(2)×cos60 ∘ Since cos ⁡ 60 ∘ = 0.5 cos60 ∘ =0.5: Φ E = 500 × 2 × 0.5 Φ E ​ =500×2×0.5 Φ E = 500 N m 2 / C Φ E ​ =500Nm 2 /C Answer: 500 N m 2 / C 500Nm 2 /C

Question 30

A point charge of + 6 μ C +6μC is placed inside a spherical surface. What is the total electric flux through the sphere?

Answer 30

Solution Using Gauss’s Law: Φ E = q ε 0 Φ E ​ = ε 0 ​ q ​ Φ E = 6 × 10 − 6 C 8.85 × 10 − 12 C 2 / N m 2 Φ E ​ = 8.85×10 −12 C 2 /Nm 2 6×10 −6 C ​ Φ E = 6.78 × 10 5 N m 2 / C Φ E ​ =6.78×10 5 Nm 2 /C Answer: 6.78 × 10 5 N m 2 / C 6.78×10 5 Nm 2 /C

Question 31

A + 10 μ C +10μC charge is placed at the center of a cube with edge length 5 cm. Find the electric flux through one face of the cube.

Answer 31

Solution By Gauss’s Law, the total flux through the cube: Φ E = q ε 0 Φ E ​ = ε 0 ​ q ​ Φ E = 10 × 10 − 6 C 8.85 × 10 − 12 Φ E ​ = 8.85×10 −12 10×10 −6 C ​ Φ E = 1.13 × 10 6 N m 2 / C Φ E ​ =1.13×10 6 Nm 2 /C Since a cube has 6 faces, the flux through one face is: Φ E face = 1.13 × 10 6 6 Φ E face ​ = 6 1.13×10 6 ​ Φ E face = 1.88 × 10 5 N m 2 / C Φ E face ​ =1.88×10 5 Nm 2 /C Answer: 1.88 × 10 5 N m 2 / C 1.88×10 5 Nm 2 /C per face

Question 32

A closed surface encloses no charge inside it. What is the total flux through the surface? A) 0 N m 2 / C 0Nm 2 /C B) 8.85 × 10 − 12 N m 2 / C 8.85×10 −12 Nm 2 /C C) 1.6 × 10 − 19 N m 2 / C 1.6×10 −19 Nm 2 /C D) 9 × 10 9 N m 2 / C 9×10 9 Nm 2 /C

Answer 32

By Gauss’s Law, if there is no charge enclosed: Φ E = q enc ε 0 = 0 Φ E ​ = ε 0 ​ q enc ​ ​ =0 Correct Answer: A) 0 N m 2 / C 0Nm 2 /C

Question 33

A + 5 μ C +5μC charge is placed inside a spherical Gaussian surface. Find the total flux through the sphere.

Answer 33

Using: Φ E = q ε 0 Φ E ​ = ε 0 ​ q ​ Φ E = 5 × 10 − 6 C 8.85 × 10 − 12 C 2 / N m 2 Φ E ​ = 8.85×10 −12 C 2 /Nm 2 5×10 −6 C ​ Φ E = 5.65 × 10 5 N m 2 / C Φ E ​ =5.65×10 5 Nm 2 /C Answer: 5.65 × 10 5 N m 2 / C 5.65×10 5 Nm 2 /C.

Question 34

A − 8 μ C −8μC charge is placed at the center of a cube. Find the flux through one face of the cube.

Answer 34

Using: Φ E = − 8 × 10 − 6 8.85 × 10 − 12 Φ E ​ = 8.85×10 −12 −8×10 −6 ​ Φ E = − 9.04 × 10 5 N m 2 / C Φ E ​ =−9.04×10 5 Nm 2 /C Flux through one face: Φ face = − 9.04 × 10 5 6 Φ face ​ = 6 −9.04×10 5 ​ Φ face = − 1.51 × 10 5 N m 2 / C Φ face ​ =−1.51×10 5 Nm 2 /C Answer: − 1.51 × 10 5 N m 2 / C −1.51×10 5 Nm 2 /C per face.

Question 35

A + 10 μ C +10μC charge is placed outside a spherical closed surface. Find the total flux through the sphere.

Answer 35

Since the charge is outside, no charge is enclosed, so: Φ E = 0 Φ E ​ =0 Answer: 0 N m 2 / C 0Nm 2 /C.

Question 36

A + 12 μ C +12μC charge is inside a spherical surface. What is the total electric flux through the surface? A) 1.36 × 10 6 N m 2 / C 1.36×10 6 Nm 2 /C B) 2.7 × 10 6 N m 2 / C 2.7×10 6 Nm 2 /C C) 1.08 × 10 6 N m 2 / C 1.08×10 6 Nm 2 /C D) 4.2 × 10 6 N m 2 / C 4.2×10 6 Nm 2 /C

Answer 36

Solution Using: Φ E = q ε 0 Φ E ​ = ε 0 ​ q ​ Φ E = 12 × 10 − 6 8.85 × 10 − 12 Φ E ​ = 8.85×10 −12 12×10 −6 ​ Φ E = 1.36 × 10 6 N m 2 / C Φ E ​ =1.36×10 6 Nm 2 /C Correct Answer: A) 1.36 × 10 6 N m 2 / C 1.36×10 6 Nm 2 /C.

Question 37

Flux Through a Cube’s Face A − 5 μ C −5μC charge is enclosed in a cube. What is the flux through one face of the cube? A) − 9.42 × 10 4 N m 2 / C −9.42×10 4 Nm 2 /C B) − 4.71 × 10 4 N m 2 / C −4.71×10 4 Nm 2 /C C) − 7.42 × 10 4 N m 2 / C −7.42×10 4 Nm 2 /C D) − 1.24 × 10 4 N m 2 / C −1.24×10 4 Nm 2 /C

Answer 37

Total flux: Φ E = − 5 × 10 − 6 8.85 × 10 − 12 Φ E ​ = 8.85×10 −12 −5×10 −6 ​ Φ E = − 5.65 × 10 5 N m 2 / C Φ E ​ =−5.65×10 5 Nm 2 /C Flux through one face: Φ face = − 5.65 × 10 5 6 Φ face ​ = 6 −5.65×10 5 ​ Φ face = − 9.42 × 10 4 N m 2 / C Φ face ​ =−9.42×10 4 Nm 2 /C Correct Answer: A) − 9.42 × 10 4 N m 2 / C −9.42×10 4 Nm 2 /C.

Question 38

A + 8 μ C +8μC charge is placed at the center of a sphere. Find the total flux through the sphere.

Answer 38

Solution: Φ E = q ε 0 Φ E ​ = ε 0 ​ q ​ Φ E = 8 × 10 − 6 8.85 × 10 − 12 Φ E ​ = 8.85×10 −12 8×10 −6 ​ Φ E = 9.04 × 10 5 N m 2 / C Φ E ​ =9.04×10 5 Nm 2 /C Answer: 9.04 × 10 5 N m 2 / C 9.04×10 5 Nm 2 /C.

Question 39

A 3 m × 2 m 3m×2m rectangular sheet is placed perpendicular to an E = 500 N / C E=500N/C field. Find the electric flux through the rectangle.

Answer 39

Solution: A = 3 × 2 = 6 m 2 A=3×2=6m 2 Φ E = ( 500 ) × ( 6 ) Φ E ​ =(500)×(6) Φ E = 3000 N m 2 / C Φ E ​ =3000Nm 2 /C Answer: 3000 N m 2 / C 3000Nm 2 /C.

Question 40

A 3 m × 4 m 3m×4m rectangular surface is placed at 30 ∘ 30 ∘ to a 700 N / C 700N/C field. Find the flux.

Answer 40

Solution: A = 3 × 4 = 12 m 2 A=3×4=12m 2 Φ E = ( 700 ) × ( 12 ) × cos ⁡ 30 ∘ Φ E ​ =(700)×(12)×cos30 ∘ Since cos ⁡ 30 ∘ = 0.866 cos30 ∘ =0.866, Φ E = 700 × 12 × 0.866 Φ E ​ =700×12×0.866 Φ E = 7274.4 N m 2 / C Φ E ​ =7274.4Nm 2 /C Answer: 7274.4 N m 2 / C 7274.4Nm 2 /C.

Question 41

A 5 m × 5 m 5m×5m square is placed parallel to a 1000 N / C 1000N/C field. Find the electric flux through it.

Answer 41

Solution: Since the square is parallel to the field, the field lines do not pass through the surface ( θ = 90 ∘ θ=90 ∘ ): Φ E = E A cos ⁡ 90 ∘ Φ E ​ =EAcos90 ∘ Since cos ⁡ 90 ∘ = 0 cos90 ∘ =0, the flux is: Φ E = 0 Φ E ​ =0 Answer: 0 N m 2 / C 0Nm 2 /C (No flux passes through the square)

Question 42

A metal plate of an electrophorus is lifted after grounding. What type of charge does it retain? A) Negative B) Positive C) Neutral D) Both positive and negative

Answer 42

Answer: B) Positive Reason: The metal plate loses negative charges when grounded, leaving it positively charged

Question 43

An electroscope is initially uncharged. A negatively charged rod is brought near the cap without touching it. What happens to the gold leaves? A) They move apart B) They collapse C) They remain the same D) They oscillate

Answer 43

Answer: A) They move apart Reason: Induced positive charge on the cap causes negative charge to move to the leaves. Since both leaves get the same charge, they repel.

Question 44

A metal plate in an electrophorus system is placed on a negatively charged insulator and grounded. The insulator carries a charge of − 8 μ C −8μC. If the metal plate has a capacitance of 20 p F 20pF (picoFarads), what charge does the plate acquire after grounding?

Answer 44

Solution Using electrostatic induction: The negative charge on the insulator induces positive charge on the metal plate’s bottom. When the metal plate is grounded, it loses its negative charges, leaving only positive charge. The charge on the metal plate is given by: Q = C V Q=CV But since capacitance is given and potential is not, we use the fact that an ideal electrophorus setup mirrors the induced charge. Thus: Q plate = + 8 μ C Q plate ​ =+8μC Final Answer: + 8 μ C +8μC (The plate gets a positive charge).

Question 45

A metal plate with + 5 μ C +5μC charge is lifted 0.02 m from a negatively charged insulator. If the electrostatic force is 0.1 N 0.1N, calculate the work done in lifting the plate.

Answer 45

Solution Work done is: W = F d W=Fd W = ( 0.1 N ) × ( 0.02 m ) W=(0.1N)×(0.02m) W = 0.002 J W=0.002J Final Answer: 2 × 10 − 3 J 2×10 −3 J or 0.002 J 0.002J.

Question 46

An electrophorus generates + 6 μ C +6μC per lift. If it is used 5 times to charge a conductor, how much total charge is transferred?

Answer 46

Q total ​ =nQ Q total = 5 × ( 6 μ C ) Q total ​ =5×(6μC) Q total = 30 μ C Q total ​ =30μC Final Answer: 30 μ C 30μC.

Question 47

Potential of an Electroscope An electroscope’s gold leaves carry a charge of 2 μ C 2μC. If the electroscope’s capacitance is 10 p F 10pF, calculate the potential of the electroscope.

Answer 47

Solution Using the formula: V = Q C V= C Q ​ V = 2 × 10 − 6 C 10 × 10 − 12 F V= 10×10 −12 F 2×10 −6 C ​ V = 2 × 10 5 V V=2×10 5 V Final Answer: 200 , 000 V 200,000V or 2 × 10 5 V 2×10 5 V.

Question 48

An electroscope initially has − 4 μ C −4μC. A + 2 μ C +2μC charge is brought into contact with the electroscope. What is the final charge?

Answer 48

Solution Since charge redistributes, the new charge is: Q final = Q initial + Q added Q final ​ =Q initial ​ +Q added ​ Q final = ( − 4 μ C ) + ( 2 μ C ) Q final ​ =(−4μC)+(2μC) Q final = − 2 μ C Q final ​ =−2μC Final Answer: − 2 μ C −2μC (Net charge remains negative).

Question 49

A charged electroscope has a gold leaf divergence of 45 ∘ 45 ∘ . If the charge on the electroscope is doubled, what happens to the leaf angle?

Answer 49

Concept The force between gold leaves follows Coulomb’s Law. Doubling the charge doubles the force and increases the angle. However, the exact relation depends on tension and leaf length. For small angles: θ new ≈ 1.4 × θ old θ new ​ ≈1.4×θ old ​ θ new = 1.4 × 45 ∘ θ new ​ =1.4×45 ∘ θ new ≈ 63 ∘ θ new ​ ≈63 ∘ Final Answer: 63 ∘ 63 ∘ (Leaves spread more).

Question 50

An electroscope is positively charged. A mystery object is brought near, and the leaves spread further. What is the object’s charge? A) Negative B) Positive C) Neutral D) Cannot be determined

Answer 50

Correct Answer: B) Positive Explanation: The object induces more positive charge on the leaves, making them repel further.

Question 51

A negatively charged rod is brought near the metal cap of a neutral electroscope, and then the cap is grounded before removing the rod. What is the final charge? A) Negative B) Positive C) Neutral D) Alternating charge

Answer 51

Correct Answer: B) Positive Explanation: Inductive charging causes positive charge to remain when the ground is removed.

Question 52

An electrophorus generates 4 μ C 4μC per lift. A Leyden jar needs 20 μ C 20μC to charge fully. How many lifts are required?

Answer 52

n= Q per lift ​ Q required ​ ​ n = 20 μ C 4 μ C n= 4μC 20μC ​ n = 5 n=5 ✅ Final Answer: 5 lifts.

Question 53

A dipole has charges + 2 μ C +2μC and − 2 μ C −2μC separated by 4 c m 4cm. Find the electric field at a point 10 cm away along the axial line.

Answer 53

Solution Given: q = 2 × 10 − 6 C q=2×10 −6 C, d = 4 c m = 0.04 m d=4cm=0.04m, r = 10 c m = 0.10 m r=10cm=0.10m, p = q d = ( 2 × 10 − 6 ) × ( 0.04 ) = 8 × 10 − 8 C ⋅ m p=qd=(2×10 −6 )×(0.04)=8×10 −8 C⋅m. Using: E axial = 1 4 π ε 0 ⋅ 2 p r 3 E axial ​ = 4πε 0 ​ 1 ​ ⋅ r 3 2p ​ E axial = 9 × 10 9 × 2 × ( 8 × 10 − 8 ) ( 0.10 ) 3 E axial ​ = (0.10) 3 9×10 9 ×2×(8×10 −8 ) ​ E axial = 1.44 × 10 4 0.001 E axial ​ = 0.001 1.44×10 4 ​ E axial = 1.44 × 10 7 N / C E axial ​ =1.44×10 7 N/C ✅ Final Answer: 1.44 × 10 7 N / C 1.44×10 7 N/C.

Question 54

Find the electric field at the same distance (10 cm) on the equatorial line.

Answer 54

Solution Using: E equatorial = 1 4 π ε 0 ⋅ p r 3 E equatorial ​ = 4πε 0 ​ 1 ​ ⋅ r 3 p ​ E equatorial = 9 × 10 9 × ( 8 × 10 − 8 ) ( 0.10 ) 3 E equatorial ​ = (0.10) 3 9×10 9 ×(8×10 −8 ) ​ E equatorial = 7.2 × 10 3 0.001 E equatorial ​ = 0.001 7.2×10 3 ​ E equatorial = 7.2 × 10 6 N / C E equatorial ​ =7.2×10 6 N/C ✅ Final Answer: 7.2 × 10 6 N / C 7.2×10 6 N/C (Half of the axial field).

Question 55

Which one of the following represents correct units for electric field strength? a. T b. N/C c. J / C d. N * m 2 * C-2

Answer 55

The correct unit for electric field strength (also called electric field intensity) is: E = F q E= q F ​ where: E E = Electric field strength (N/C), F F = Force (N), q q = Charge (C). Thus, the SI unit of electric field strength is Newton per Coulomb (N/C). ✅ Correct Answer: (b) N/C Explanation of Other Options (a) T (Tesla) ❌ Tesla is the unit of magnetic field strength, not electric field. (c) J/C (Joule per Coulomb) ❌ This is the unit of electric potential (Voltage, V). (d) N·m²·C − 2 −2 (Newton meter squared per Coulomb squared) ❌ This is the unit of Coulomb’s constant (k) in Coulomb's law.

Question 56

The flow of charge per unit time defines a. power. b. current. c. voltage. d. resistance.

Answer 56

The correct answer is: ✅ (b) Current Explanation The flow of charge per unit time is called electric current and is given by the formula: I = Q t I= t Q ​ where: I I = Current (Ampere, A) Q Q = Charge (Coulomb, C) t t = Time (seconds, s) Explanation of Other Options (a) Power ❌ Power is the rate at which energy is transferred: P = V I P=VI Unit: Watts (W) (c) Voltage ❌ Voltage (Potential Difference) is the work done per unit charge: V = W Q V= Q W ​ Unit: Volts (V) (d) Resistance ❌ Resistance opposes current flow and is given by: R = V I R= I V ​ Unit: Ohms (Ω) Key Takeaway Electric current is the rate at which electric charge flows through a conductor.

Question 57

Answer 57

Solution to the Electric Field Direction at Point P Step 1: Understanding the Setup There are two positive charges: Charge Q Q on the left. Charge 2 Q 2Q on the right. Point P is equidistant from both charges. Step 2: Direction of the Electric Field The electric field always points away from positive charges. At Point P, the field due to: Charge Q Q (left charge) points away from Q Q, meaning it is directed toward the right. Charge 2 Q 2Q (right charge) also points away, meaning it is directed toward the left. Step 3: Comparing Field Strengths The magnitude of the electric field due to a charge is: E = k Q r 2 E= r 2 kQ ​ where k k is Coulomb's constant, and r r is the distance from the charge. Since the charge on the right is 2 Q 2Q (double the left charge), its field is twice as strong as that of the left charge at P. The net electric field will be closer in direction to the stronger field, meaning it will point slightly away from the stronger charge (2Q), but not directly horizontal. Step 4: Choosing the Correct Vector The correct answer is the vector pointing slightly away from the stronger charge (2Q), meaning slightly to the left and upward. Looking at the options in the diagram, option (c) best represents this direction. ✅ Final Answer: Option C

Question 57

Answer 57

Solution: Electric Potential at Point P The electric potential at a point due to a charge is given by: V = k Q r V= r kQ ​ where: k = 9.0 × 10 9   N \cdotp m 2 / C 2 k=9.0×10 9 N\cdotpm 2 /C 2 (Coulomb's constant), Q Q is the charge (Coulombs), r r is the distance from the charge to the point. Step 1: Calculate the Potential Due to Each Charge at P Point P is midway between the two charges, so: Distance from Q 1 Q 1 ​ to P: r 1 = 2.0 r 1 ​ =2.0 m. Distance from Q 2 Q 2 ​ to P: r 2 = 2.0 r 2 ​ =2.0 m. Potential Due to Q 1 = 6.0 × 10 − 6 C Q 1 ​ =6.0×10 −6 C V 1 = ( 9.0 × 10 9 ) × ( 6.0 × 10 − 6 ) 2.0 V 1 ​ = 2.0 (9.0×10 9 )×(6.0×10 −6 ) ​ V 1 = 54 × 10 3 2 V 1 ​ = 2 54×10 3 ​ V 1 = 2.7 × 10 4 V V 1 ​ =2.7×10 4 V Potential Due to Q 2 = − 3.0 × 10 − 6 C Q 2 ​ =−3.0×10 −6 C V 2 = ( 9.0 × 10 9 ) × ( − 3.0 × 10 − 6 ) 2.0 V 2 ​ = 2.0 (9.0×10 9 )×(−3.0×10 −6 ) ​ V 2 = − 27 × 10 3 2 V 2 ​ = 2 −27×10 3 ​ V 2 = − 1.35 × 10 4 V V 2 ​ =−1.35×10 4 V Step 2: Find the Total Potential at P V total = V 1 + V 2 V total ​ =V 1 ​ +V 2 ​ V total = ( 2.7 × 10 4 ) + ( − 1.35 × 10 4 ) V total ​ =(2.7×10 4 )+(−1.35×10 4 ) V total = 1.35 × 10 4 V V total ​ =1.35×10 4 V Final Answer: 1.4 × 10 4 V 1.4×10 4 V ✅ Correct Option: (c) 1.4 × 10 4 V 1.4×10 4 V

Question 58

Answer 58

Solution to Question 5: Identifying Charge Polarity from Electric Field Lines Step 1: Understanding Electric Field Line Behavior Electric field lines always point away from positive charges. Electric field lines always point toward negative charges. The density of field lines indicates the relative strength of the charge. Step 2: Analyzing the Given Diagram The field lines appear to originate from Charge R and terminate at Charge L. This means Charge R is positive (since field lines radiate outward). Charge L is negative (since field lines terminate on it). ✅ Correct Answer: (c) Charge L is Negative, Charge R is Positive.

Question 59

The electric field 2.0 m from a point charge has a magnitude of 8. 0 x 10 4 N/C. What is the strength of the electric field at a distance of 4.0 m? a. 2.0 x 104 N/C b. 4.0 x 104 N/C c. 1.6 x 10 5 N/C d. 3.2 x 10 5 N/C

Answer 59

Solution: Electric Field at a Different Distance The electric field ( E E) due to a point charge follows the inverse square law: E∝1/r^2 This means that when the distance is doubled, the field strength reduces by a factor of: (r1/r2)^2 Step 1: Identify Given Values Initial distance: r 1 = 2.0 r 1 ​ =2.0 m Initial field strength: E 1 = 8.0 × 10 4 E 1 ​ =8.0×10 4 N/C New distance: r 2 = 4.0 r 2 ​ =4.0 m New field strength: E 2 = ? E 2 ​ =? Step 2: Apply Inverse Square Law E 2 = E 1 × ( r 1 r 2 ) 2 E 2 ​ =E 1 ​ ×( r 2 ​ r 1 ​ ​ ) 2 E 2 = ( 8.0 × 10 4 ) × ( 2.0 4.0 ) 2 E 2 ​ =(8.0×10 4 )×( 4.0 2.0 ​ ) 2 E 2 = ( 8.0 × 10 4 ) × ( 1 2 ) 2 E 2 ​ =(8.0×10 4 )×( 2 1 ​ ) 2 E 2 = ( 8.0 × 10 4 ) × 1 4 E 2 ​ =(8.0×10 4 )× 4 1 ​ E 2 = 2.0 × 10 4  N/C E 2 ​ =2.0×10 4  N/C ✅ Correct Answer: (a) 2.0 × 10 4 2.0×10 4 N/C

Question 59

When a charge is accelerated through a potential difference of 500 V, its kinetic energy increases from 2.0 x 10 -5 J to 6. 0 x 10 -5 J. What is the magnitude of the charge? a. 4.0 x 10 -8 C b. 8.0 x 10 -8 C c. 1. 2 x 10 -7 C d. 1. 6 x 10 -7 C

Answer 59

Solution: Finding the Charge Magnitude The work done by an electric field on a charge is given by: W = q V W=qV where: W W = Change in kinetic energy (J), q q = Charge magnitude (C), V V = Potential difference (V). Step 1: Identify Given Values Initial kinetic energy: K E initial = 2.0 × 10 − 5 KE initial ​ =2.0×10 −5 J Final kinetic energy: K E final = 6.0 × 10 − 5 KE final ​ =6.0×10 −5 J Potential difference: V = 500 V=500 V Change in kinetic energy (Work done by the field): W = K E final − K E initial W=KE final ​ −KE initial ​ W = ( 6.0 × 10 − 5 ) − ( 2.0 × 10 − 5 ) W=(6.0×10 −5 )−(2.0×10 −5 ) W = 4.0 × 10 − 5  J W=4.0×10 −5  J Step 2: Solve for Charge q q Using: q = W V q= V W ​ q = 4.0 × 10 − 5 J 500 V q= 500V 4.0×10 −5 J ​ q = 8.0 × 10 − 8 C q=8.0×10 −8 C ✅ Correct Answer: (b) 8.0 × 10 − 8 C 8.0×10 −8 C

Question 60

A negative charge in an electric field experiences a force accelerating it due south. What is the direction of the electric field? a. east b. west c. north d. south

Answer 60

Solution: Determining the Direction of the Electric Field The force on a charge in an electric field is given by: F = q E F=qE where: F F is the force on the charge. q q is the charge (negative in this case). E E is the electric field direction. Step 1: Understanding the Effect on a Negative Charge A positive charge moves in the direction of the electric field. A negative charge moves opposite to the electric field. Since the negative charge accelerates due south, the force F F is southward. But, because the charge is negative, the electric field must be in the opposite direction. Thus, the electric field is directed north. ✅ Correct Answer: (c) North

Question 61

Answer 61

Solution: Finding the Unknown Charge Using Coulomb’s Law Coulomb’s Law is given by: F= k∣q1q2∣/r^2 where: F=0.35 N (force), k=9.0×10 ^9 Nm²/C² (Coulomb’s constant), q 1=−2.3×10 ^−6 C (known charge), q 2=? (unknown charge), r=0.20 m (distance between charges). Step 1: Solve for q 2 q 2 ​ Rearranging Coulomb’s law to solve for q 2 q 2 ​ : q 2 = F r 2 k q 1 q 2 ​ = kq 1 ​ Fr 2 ​ Substituting values: q 2 = ( 0.35 ) ( 0.20 ) 2 ( 9.0 × 10 9 ) ( 2.3 × 10 − 6 ) q 2 ​ = (9.0×10 9 )(2.3×10 −6 ) (0.35)(0.20) 2 ​ q 2 = ( 0.35 ) ( 0.04 ) ( 9.0 × 10 9 ) ( 2.3 × 10 − 6 ) q 2 ​ = (9.0×10 9 )(2.3×10 −6 ) (0.35)(0.04) ​ q 2 = 0.014 2.07 × 10 4 q 2 ​ = 2.07×10 4 0.014 ​ q 2 = 6.8 × 10 − 7 C q 2 ​ =6.8×10 −7 C Step 2: Determine the Polarity of q 2 q 2 ​ The problem states that the force is repulsive. Since like charges repel, and q 1 = − 2.3 × 10 − 6 C q 1 ​ =−2.3×10 −6 C (negative), q 2 q 2 ​ must also be negative. ✅ Final Answer: (a) 6.8 × 10 − 7 C 6.8×10 −7 C, Negative.

Question 62

Answer 62

Solution: Electric Field at Midpoint (Point P) The electric field ( E E) due to a point charge is given by: E = k ∣ q ∣ r 2 E= r 2 k∣q∣ ​ where: k = 9.0 × 10 9 k=9.0×10 9 Nm²/C² (Coulomb’s constant), q q is the charge, r r is the distance from the charge to the point. Step 1: Determine Individual Fields at P The midpoint P P is 1.5 m from each charge. The two charges are opposite in sign, so their electric fields add up (instead of canceling). The electric field due to a positive charge points away from the charge. The electric field due to a negative charge points toward the charge. Since both fields at P P point toward the negative charge (rightward), we add their magnitudes. Field Due to q 1 = 2.5 × 10 − 6 C q 1 ​ =2.5×10 −6 C E 1 = ( 9.0 × 10 9 ) ( 2.5 × 10 − 6 ) ( 1.5 ) 2 E 1 ​ = (1.5) 2 (9.0×10 9 )(2.5×10 −6 ) ​ E 1 = ( 2.25 × 10 4 ) 2.25 E 1 ​ = 2.25 (2.25×10 4 ) ​ E 1 = 1.0 × 10 4  N/C E 1 ​ =1.0×10 4  N/C Field Due to q 2 = − 5.0 × 10 − 6 C q 2 ​ =−5.0×10 −6 C E 2 = ( 9.0 × 10 9 ) ( 5.0 × 10 − 6 ) ( 1.5 ) 2 E 2 ​ = (1.5) 2 (9.0×10 9 )(5.0×10 −6 ) ​ E 2 = ( 4.5 × 10 4 ) 2.25 E 2 ​ = 2.25 (4.5×10 4 ) ​ E 2 = 2.0 × 10 4  N/C E 2 ​ =2.0×10 4  N/C Step 2: Add the Fields Since both fields point in the same direction (rightward): E total = E 1 + E 2 E total ​ =E 1 ​ +E 2 ​ E total = ( 1.0 × 10 4 ) + ( 2.0 × 10 4 ) E total ​ =(1.0×10 4 )+(2.0×10 4 ) E total = 3.0 × 10 4  N/C E total ​ =3.0×10 4  N/C ✅ Final Answer: (d) 3.0 × 10 4 3.0×10 4 N/C.

Question 63

Answer 63

Solution: Work Done in Moving a Charge The work done in moving a charge in an electric field is given by the change in electric potential energy: W = U f − U i W=U f ​ −U i ​ Since electric potential energy between two point charges is: U = k q 1 q 2 r U= r kq 1 ​ q 2 ​ ​ where: k = 9.0 × 10 9 k=9.0×10 9 Nm²/C² (Coulomb’s constant), q 1 = 4.0 × 10 − 9 C q 1 ​ =4.0×10 −9 C (moving charge), q 2 = 6.0 × 10 − 8 C q 2 ​ =6.0×10 −8 C (stationary charge), r i = 3.0 r i ​ =3.0 m (initial distance), r f = 0.50 r f ​ =0.50 m (final distance). Step 1: Compute Initial Potential Energy U i = ( 9.0 × 10 9 ) ( 4.0 × 10 − 9 ) ( 6.0 × 10 − 8 ) 3.0 U i ​ = 3.0 (9.0×10 9 )(4.0×10 −9 )(6.0×10 −8 ) ​ U i = 2.16 × 10 − 6 3.0 U i ​ = 3.0 2.16×10 −6 ​ U i = 7.2 × 10 − 7 J U i ​ =7.2×10 −7 J Step 2: Compute Final Potential Energy U f = ( 9.0 × 10 9 ) ( 4.0 × 10 − 9 ) ( 6.0 × 10 − 8 ) 0.50 U f ​ = 0.50 (9.0×10 9 )(4.0×10 −9 )(6.0×10 −8 ) ​ U f = 2.16 × 10 − 6 0.50 U f ​ = 0.50 2.16×10 −6 ​ U f = 4.32 × 10 − 6 J U f ​ =4.32×10 −6 J Step 3: Compute Work Done W = U f − U i W=U f ​ −U i ​ W = ( 4.32 × 10 − 6 ) − ( 7.2 × 10 − 7 ) W=(4.32×10 −6 )−(7.2×10 −7 ) W = 3.6 × 10 − 6 J W=3.6×10 −6 J ✅ Final Answer: (c) 3.6 × 10 − 6 J 3.6×10 −6 J.

Question 64

Answer 64

Solution: Comparing Speed and Kinetic Energy of an Electron and Proton in an Electric Field Step 1: Understanding the Setup Two parallel plates are 4.0 × 10⁻² m apart. Potential difference between plates: 1000 V. An electron starts from the negative plate, and a proton starts from the positive plate. Both are accelerated toward the opposite plate by the electric field. Step 2: Determine the Kinetic Energy The work done by the electric field on a charge is converted into kinetic energy: K E = q V KE=qV where: q = 1.6 × 10 − 19 C q=1.6×10 −19 C (for both electron and proton), V = 1000 V V=1000V. Since both the electron and proton have the same charge magnitude and experience the same potential difference, the kinetic energy they gain is: K E electron = K E proton = ( 1.6 × 10 − 19 C ) ( 1000 V ) KE electron ​ =KE proton ​ =(1.6×10 −19 C)(1000V) K E = 1.6 × 10 − 16 J KE=1.6×10 −16 J Thus, both the electron and proton have the same kinetic energy when they reach the opposite plate. ✅ Kinetic Energy: Same Step 3: Determine the Speed of Electron and Proton Kinetic energy is also given by: K E = 1 2 m v 2 KE= 2 1 ​ mv 2 Solving for velocity: v = 2 K E m v= m 2KE ​ ​ The mass of an electron is: m e = 9.11 × 10 − 31  kg m e ​ =9.11×10 −31  kg The mass of a proton is: m p = 1.67 × 10 − 27  kg m p ​ =1.67×10 −27  kg Since m p ≫ m e m p ​ ≫m e ​ , the electron, having much less mass, will have a much greater velocity than the proton. ✅ Speed: Different Final Answer The kinetic energy is the same, but the speed is different because the electron is much lighter and moves much faster than the proton. ✅ Correct Answer: (c) Different speed, same kinetic energy.

Question 65

Answer 65

Solution: Electric Field at Points r r and s s Step 1: Understanding the Electric Field Around a Positive Charge The electric field due to a positive charge ( Q Q) points radially outward. The electric field magnitude at a distance r r from a point charge is given by: E = k Q r 2 E= r 2 kQ ​ where: k = 9.0 × 10 9 k=9.0×10 9 Nm²/C² (Coulomb’s constant), Q Q = charge magnitude, r r = distance from the charge. Step 2: Comparing the Magnitude of the Electric Field at r r and s s If r r and s s are at different distances from Q Q, then their electric field magnitudes will be different because electric field strength follows an inverse square law. The closer point will experience a stronger electric field. Thus, the magnitude of the field at r r and s s is different. ✅ Magnitude: Different Step 3: Determining the Direction of the Electric Field Since Q Q is positive, the electric field radiates outward. At both r r and s s, the field direction will be away from Q Q. ✅ Direction: Away from Q Q Final Answer The magnitude of the field is different, and the direction of the field is away from Q Q. ✅ Correct Answer: (c) Different magnitude, away from Q Q.

Question 66

Answer 66

Solution: Net Force on the 2.0 × 10 − 6 C 2.0×10 −6 C Charge Step 1: Understanding the Setup We have: A + 8.0 × 10 − 6 C +8.0×10 −6 C charge on the left. A − 5.0 × 10 − 6 C −5.0×10 −6 C charge on the right. A + 2.0 × 10 − 6 C +2.0×10 −6 C charge located midway between them. The distances between the charges are 2.0 m. The force between two point charges is given by Coulomb’s Law: F = k ∣ q 1 q 2 ∣ r 2 F=k r 2 ∣q 1 ​ q 2 ​ ∣ ​ where: k = 9.0 × 10 9 k=9.0×10 9 Nm²/C², q 1 , q 2 q 1 ​ ,q 2 ​ are the charges, r r is the separation between charges. Step 2: Compute Forces on the 2.0 × 10 − 6 C 2.0×10 −6 C Charge Force Due to + 8.0 × 10 − 6 C +8.0×10 −6 C (on the left) F 1 = ( 9.0 × 10 9 ) ( 8.0 × 10 − 6 ) ( 2.0 × 10 − 6 ) ( 2.0 ) 2 F 1 ​ = (2.0) 2 (9.0×10 9 )(8.0×10 −6 )(2.0×10 −6 ) ​ F 1 = 1.44 × 10 − 1 4 F 1 ​ = 4 1.44×10 −1 ​ F 1 = 3.6 × 10 − 2 N F 1 ​ =3.6×10 −2 N ✅ Direction: Since both charges are positive, the repulsive force pushes the 2.0 × 10 − 6 C 2.0×10 −6 C charge to the right. Force Due to − 5.0 × 10 − 6 C −5.0×10 −6 C (on the right) F 2 = ( 9.0 × 10 9 ) ( 5.0 × 10 − 6 ) ( 2.0 × 10 − 6 ) ( 2.0 ) 2 F 2 ​ = (2.0) 2 (9.0×10 9 )(5.0×10 −6 )(2.0×10 −6 ) ​ F 2 = 9.0 × 10 − 2 4 F 2 ​ = 4 9.0×10 −2 ​ F 2 = 2.25 × 10 − 2 N F 2 ​ =2.25×10 −2 N ✅ Direction: Since one charge is positive and the other is negative, the attractive force pulls the 2.0 × 10 − 6 C 2.0×10 −6 C charge to the right. Step 3: Find the Net Force Both forces F 1 F 1 ​ and F 2 F 2 ​ act in the same direction (right), so the total force is: F net = F 1 + F 2 F net ​ =F 1 ​ +F 2 ​ F net = ( 3.6 × 10 − 2 ) + ( 2.25 × 10 − 2 ) F net ​ =(3.6×10 −2 )+(2.25×10 −2 ) F net = 5.85 × 10 − 2 N F net ​ =5.85×10 −2 N Rounding to 5.9 × 10 − 2 N 5.9×10 −2 N. ✅ Final Answer: (d) 5.9 × 10 − 2 N 5.9×10 −2 N towards the right.

Question 67

What is the electric potential energy of an electron located 5. 3 x 10 -11 m from the proton in a hydrogen atom? a. -8. 2 x 10 -8 J b. -4.3 x 10 -18 J c. -2.2 x 10 -18 J d. -1.6 x 10 -19 J

Answer 67

Solution: Electric Potential Energy of an Electron in a Hydrogen Atom The electric potential energy ( U U) between two point charges is given by Coulomb’s Law: U = k q 1 q 2 r U= r kq 1 ​ q 2 ​ ​ where: k = 9.0 × 10 9 k=9.0×10 9 Nm²/C² (Coulomb’s constant), q 1 = q 2 = e = 1.6 × 10 − 19 C q 1 ​ =q 2 ​ =e=1.6×10 −19 C (charge of electron and proton), r = 5.3 × 10 − 11 r=5.3×10 −11 m (Bohr radius in hydrogen atom). Step 1: Substitute the Given Values U = ( 9.0 × 10 9 ) ( 1.6 × 10 − 19 ) ( 1.6 × 10 − 19 ) ( 5.3 × 10 − 11 ) U= (5.3×10 −11 ) (9.0×10 9 )(1.6×10 −19 )(1.6×10 −19 ) ​ U = ( 9.0 × 10 9 ) ( 2.56 × 10 − 38 ) 5.3 × 10 − 11 U= 5.3×10 −11 (9.0×10 9 )(2.56×10 −38 ) ​ U = 2.304 × 10 − 28 5.3 × 10 − 11 U= 5.3×10 −11 2.304×10 −28 ​ U = − 4.35 × 10 − 18 J U=−4.35×10 −18 J Rounding to − 4.3 × 10 − 18 −4.3×10 −18 J. Step 2: Consider the Sign Since the electron and proton attract, the potential energy is negative. ✅ Final Answer: (b) − 4.3 × 10 − 18 −4.3×10 −18 J.

Question 68

Answer 68

Solution: Potential Difference Between Point P and One of the Plates Step 1: Understanding the Problem Two parallel plates are separated by 0.028 m. The potential difference between the plates is 80 V. Point P is located midway between the plates, meaning it is at a distance of: 0.028 2 = 0.014  m 2 0.028 ​ =0.014 m from either plate. The electric potential varies linearly between the plates. Step 2: Calculate the Potential at Point P The electric field between parallel plates is uniform and given by: E = V d E= d V ​ where: V = 80 V V=80V (total potential difference), d = 0.028 m d=0.028m (distance between plates). E = 80 0.028 E= 0.028 80 ​ E = 2857.14  V/m E=2857.14 V/m Since the electric field is uniform, the potential at a distance x x from the 0 V plate is: V P = E × x V P ​ =E×x At x = 0.014 m x=0.014m: V P = 2857.14 × 0.014 V P ​ =2857.14×0.014 V P = 40 V V P ​ =40V Thus, point P is at 40 V. Step 3: Find the Potential Difference Between P and One Plate The potential difference between P and the 0V plate is 40 V. The potential difference between P and the 80V plate is also 40 V. ✅ Final Answer: (b) 40 V.

Question 69

Answer 69

Solution: Finding the Point Where Electric Field is Zero Step 1: Understanding the Concept The electric field at any point due to a charge is given by: E = k Q r 2 E= r 2 kQ ​ where: k k is Coulomb’s constant, Q Q is the charge magnitude, r r is the distance from the charge. Two equal positive charges create electric fields that point away from each charge. The electric field cancels out at a point where the fields due to both charges are equal in magnitude but opposite in direction. Step 2: Identify the Zero Field Location Inside the region between the two charges: In this case, the electric fields add up instead of canceling because both fields point outward from the positive charges. Thus, no zero-field point exists between the charges. Outside the region, on the left or right side of both charges: The zero-field point must be on the outer side of one of the charges, where the two electric fields oppose each other. Since the charges are equal, the zero-field point is at an equal distance from both charges on the left or right side. This means point 1 or point 4 could be the correct answer. However, the zero-field point will be closer to the midpoint outside the charge pair, meaning point 1 is the correct location. ✅ Final Answer: (a) Location 1.

Question 70

An electron is positioned in an electric field. The force on the electron due to the electric field is equal to the force of gravity on the electron. What is the magnitude of this electric field? a. 8.93 x 10 -30 N/C b. 5.69 x 10 -12 N/C c. 5.58 x 10 -11 N/C d. 1.44 x 10 -9 N/C

Answer 70

Solution: Finding the Electric Field Magnitude Step 1: Understanding the Forces Acting on the Electron The problem states that the electric force acting on the electron is equal to the gravitational force acting on it. Electric force: F e = q E F e ​ =qE where: q = 1.6 × 10 − 19 C q=1.6×10 −19 C (charge of an electron), E E is the electric field (to be found). Gravitational force: F g = m g F g ​ =mg where: m = 9.11 × 10 − 31 k g m=9.11×10 −31 kg (mass of an electron), g = 9.81 g=9.81 m/s² (acceleration due to gravity). Since the forces are equal: q E = m g qE=mg Step 2: Solve for E E E = m g q E= q mg ​ Substituting values: E = ( 9.11 × 10 − 31 ) ( 9.81 ) 1.6 × 10 − 19 E= 1.6×10 −19 (9.11×10 −31 )(9.81) ​ E = 8.93 × 10 − 30 1.6 × 10 − 19 E= 1.6×10 −19 8.93×10 −30 ​ E = 5.58 × 10 − 11  N/C E=5.58×10 −11  N/C Final Answer ✅ (c) 5.58 × 10 − 11 5.58×10 −11 N/C

Question 71

Answer 71

Solution: Electric Field Magnitude at Point P For parallel plates, the electric field ( E E) between them is uniform and given by: E = V d E= d V ​ where: V = 850 V=850 V (potential difference between the plates), d = 2.5 × 10 − 2 d=2.5×10 −2 m (separation between the plates). Step 1: Calculate the Electric Field E = 850 2.5 × 10 − 2 E= 2.5×10 −2 850 ​ E = 850 0.025 E= 0.025 850 ​ E = 3.4 × 10 4  V/m E=3.4×10 4  V/m Final Answer ✅ (c) 3.4 × 10 4 3.4×10 4 V/m

Question 72

. A particle with a charge of 2. 4 x 10 -5 C is accelerated from rest through a potential difference of 6.2 x 10 4 V. If the final speed of this particle is 9.3 x 103 m/s, what is the mass of the particle? a. 7.7 x 10-10 kg b. 5.2 x 10-9 kg c. 3.4 x 10-8 kg d. 1.5 x 10-1 kg

Answer 72

Solution: Finding the Mass of the Particle Step 1: Use Energy Conservation The work done by the electric field on the particle is converted into kinetic energy: q V = 1 2 m v 2 qV= 2 1 ​ mv 2 where: q = 2.4 × 10 − 5 C q=2.4×10 −5 C (charge of the particle), V = 6.2 × 10 4 V V=6.2×10 4 V (potential difference), v = 9.3 × 10 3 v=9.3×10 3 m/s (final velocity), m m = mass of the particle (to be found). Step 2: Solve for m m Rearrange the equation: m = 2 q V v 2 m= v 2 2qV ​ Substituting values: m = 2 ( 2.4 × 10 − 5 ) ( 6.2 × 10 4 ) ( 9.3 × 10 3 ) 2 m= (9.3×10 3 ) 2 2(2.4×10 −5 )(6.2×10 4 ) ​ m = 2.976 × 10 0 8.649 × 10 7 m= 8.649×10 7 2.976×10 0 ​ m = 3.44 × 10 − 8  kg m=3.44×10 −8  kg Final Answer ✅ (c) 3.4 × 10 − 8 3.4×10 −8 kg.

Question 73

Answer 73

Solution: Electric Potential at Point P Due to Two Charges The electric potential ( V V) at a point due to a point charge is given by: V = k Q r V= r kQ ​ where: k = 9.0 × 10 9 k=9.0×10 9 Nm²/C² (Coulomb’s constant), Q = 3.0 × 10 − 6 C Q=3.0×10 −6 C (each charge), r = 5.0 r=5.0 m (distance from each charge to point P). Step 1: Calculate Potential Due to One Charge Since electric potential is a scalar quantity, we can directly add the potentials from both charges. For one charge: V 1 = ( 9.0 × 10 9 ) ( 3.0 × 10 − 6 ) 5.0 V 1 ​ = 5.0 (9.0×10 9 )(3.0×10 −6 ) ​ V 1 = 27.0 × 10 3 5.0 V 1 ​ = 5.0 27.0×10 3 ​ V 1 = 5.4 × 10 3 V V 1 ​ =5.4×10 3 V Since there are two identical charges, the total potential at point P is: V total = V 1 + V 2 V total ​ =V 1 ​ +V 2 ​ V total = ( 5.4 × 10 3 ) + ( 5.4 × 10 3 ) V total ​ =(5.4×10 3 )+(5.4×10 3 ) V total = 1.1 × 10 4 V V total ​ =1.1×10 4 V Final Answer ✅ (d) 1.1 × 10 4 1.1×10 4 V.

Question 74

An electron experiences an electric force of 1.8 x 10-11 N at a distance of 5.0 x 10-9 m from the nucleus of an ion. The electron is moved farther away, to a distance of 2.0 x 10 -8 m from the ion. What is the new electric force on the electron? a. 1.1 x 10-12 N b. 4.5 x 10-12 N c. 7. 2 x 10-11 N d. 2.9 x 10-10 N

Answer 74

Solution: Finding the New Electric Force Using Coulomb’s Law The electric force between two charges follows Coulomb’s Law: F = k ∣ q 1 q 2 ∣ r 2 F=k r 2 ∣q 1 ​ q 2 ​ ∣ ​ Since the charge of the electron and the ion remain the same, we use the inverse square law property: F 2 = F 1 × ( r 1 r 2 ) 2 F 2 ​ =F 1 ​ ×( r 2 ​ r 1 ​ ​ ) 2 where: F 1 = 1.8 × 10 − 11 F 1 ​ =1.8×10 −11 N (initial force), r 1 = 5.0 × 10 − 9 r 1 ​ =5.0×10 −9 m (initial distance), r 2 = 2.0 × 10 − 8 r 2 ​ =2.0×10 −8 m (new distance). Step 1: Apply the Inverse Square Law F 2 = ( 1.8 × 10 − 11 ) × ( 5.0 × 10 − 9 2.0 × 10 − 8 ) 2 F 2 ​ =(1.8×10 −11 )×( 2.0×10 −8 5.0×10 −9 ​ ) 2 F 2 = ( 1.8 × 10 − 11 ) × ( 5 20 ) 2 F 2 ​ =(1.8×10 −11 )×( 20 5 ​ ) 2 F 2 = ( 1.8 × 10 − 11 ) × ( 1 4 ) F 2 ​ =(1.8×10 −11 )×( 4 1 ​ ) F 2 = 1.8 × 10 − 11 4 F 2 ​ = 4 1.8×10 −11 ​ F 2 = 4.5 × 10 − 12 N F 2 ​ =4.5×10 −12 N Final Answer ✅ (b) 4.5 × 10 − 12 4.5×10 −12 N.

Question 75

Answer 75

Solution: Identifying the Electric Field of Two Opposite Point Charges (Dipole) Step 1: Understanding the Electric Field of Opposite Charges Electric field lines always point away from positive charges and toward negative charges. For two equal but opposite charges (a dipole): The field lines originate from the positive charge and terminate at the negative charge. The field pattern shows curved lines connecting the two charges, forming a distinctive dipole shape. Step 2: Analyze the Given Options (a) & (b): Both show field lines radiating outward, which occurs when both charges are positive or negative (not opposite charges). (c): Shows correct behavior where field lines originate from the positive charge and curve toward the negative charge. (d): Shows closed-loop magnetic field lines, which is incorrect for electric fields. ✅ Final Answer: (c) – This correctly represents the electric field of two equal but opposite charges (dipole).

Question 76

Answer 76

Solution: Finding the Electric Field Magnitude at Point P We are given two point charges: 5.0 × 10 − 6 C 5.0×10 −6 C located 3.0 m to the left of P. 8.0 × 10 − 6 C 8.0×10 −6 C located 3.0 m below P. We need to calculate the electric field at point P due to both charges and find the net electric field. Step 1: Use the Electric Field Formula The electric field ( E E) due to a point charge is: E = k Q r 2 E= r 2 kQ ​ where: k = 9.0 × 10 9 k=9.0×10 9 Nm²/C² (Coulomb’s constant), Q Q is the charge magnitude, r r is the distance from the charge to P. Step 2: Calculate E 1 E 1 ​ (Due to 5.0 × 10 − 6 C 5.0×10 −6 C) E 1 = ( 9.0 × 10 9 ) ( 5.0 × 10 − 6 ) ( 3.0 ) 2 E 1 ​ = (3.0) 2 (9.0×10 9 )(5.0×10 −6 ) ​ E 1 = ( 45.0 × 10 3 ) 9 E 1 ​ = 9 (45.0×10 3 ) ​ E 1 = 5.0 × 10 3  N/C E 1 ​ =5.0×10 3  N/C Direction: Since the charge is positive, the field at P points away, meaning it points to the right. Step 3: Calculate E 2 E 2 ​ (Due to 8.0 × 10 − 6 C 8.0×10 −6 C) E 2 = ( 9.0 × 10 9 ) ( 8.0 × 10 − 6 ) ( 3.0 ) 2 E 2 ​ = (3.0) 2 (9.0×10 9 )(8.0×10 −6 ) ​ E 2 = ( 72.0 × 10 3 ) 9 E 2 ​ = 9 (72.0×10 3 ) ​ E 2 = 8.0 × 10 3  N/C E 2 ​ =8.0×10 3  N/C Direction: Since the charge is positive, the field at P points away, meaning it points upward. Step 4: Find the Net Electric Field Since E 1 E 1 ​ is horizontal (rightward) and E 2 E 2 ​ is vertical (upward), they form a right triangle. The resultant field ( E net E net ​ ) is found using the Pythagorean theorem: E net = E 1 2 + E 2 2 E net ​ = E 1 2 ​ +E 2 2 ​ ​ E net = ( 5.0 × 10 3 ) 2 + ( 8.0 × 10 3 ) 2 E net ​ = (5.0×10 3 ) 2 +(8.0×10 3 ) 2 ​ E net = ( 25.0 × 10 6 ) + ( 64.0 × 10 6 ) E net ​ = (25.0×10 6 )+(64.0×10 6 ) ​ E net = 89.0 × 10 6 E net ​ = 89.0×10 6 ​ E net ≈ 9.4 × 10 3  N/C E net ​ ≈9.4×10 3  N/C Final Answer ✅ (c) 9.4 × 10 3 9.4×10 3 N/C.

Question 77

Answer 77

Solution: Electrostatic Force on the Electron in the Electric Field Step 1: Understanding the Setup The electron is traveling horizontally between two parallel plates. The top plate is positively charged and the bottom plate is negatively charged, meaning the electric field points downward (from + + to − −). Since the electron is negatively charged, it experiences a force opposite to the electric field direction, which means the force is directed upward. Step 2: Determine if the Force is Constant The electric field between parallel plates is uniform, meaning the force on the electron remains constant throughout its motion. Since the charge of the electron is constant and the field strength does not vary, the magnitude of the force is constant. Step 3: Determine the Direction of the Force The electric field is downward, so the electron experiences an upward force (opposite to the field). Final Answer ✅ (c) Constant force, Upward direction.

Question 78

Answer 78

Solution: Finding the Maximum Speed of the Proton Step 1: Use Energy Conservation The work done by the electric field on the proton is converted into kinetic energy: q V = 1 2 m v 2 qV= 2 1 ​ mv 2 where: q = 1.6 × 10 − 19 C q=1.6×10 −19 C (charge of a proton), V = 300 V V=300V (potential difference), m = 1.67 × 10 − 27 m=1.67×10 −27 kg (mass of a proton), v v = final velocity (to be determined). Step 2: Solve for v v Rearrange the equation: v = 2 q V m v= m 2qV ​ ​ Substituting values: v = 2 ( 1.6 × 10 − 19 ) ( 300 ) 1.67 × 10 − 27 v= 1.67×10 −27 2(1.6×10 −19 )(300) ​ ​ v = 9.6 × 10 − 17 1.67 × 10 − 27 v= 1.67×10 −27 9.6×10 −17 ​ ​ v = 5.75 × 10 10 v= 5.75×10 10 ​ v ≈ 2.4 × 10 5  m/s v≈2.4×10 5  m/s Final Answer ✅ (c) 2.4 × 10 5 2.4×10 5 m/s.

Question 79

Answer 79

Solution: Identifying the Correct Electric Field Between Parallel Plates Step 1: Understanding the Electric Field Between Parallel Plates The electric field between two parallel plates is uniform and directed from the positive plate to the negative plate. Inside the plates, the field lines should be: Straight and parallel (indicating a uniform field). Equally spaced (showing uniform strength). Pointing from the positively charged plate to the negatively charged plate. Outside the plates, fringe effects cause the field lines to curve slightly. Step 2: Analyzing the Diagrams (a) & (c): These show straight, uniform, and parallel field lines inside the plates, correctly indicating a uniform electric field. (b) & (d): These show curved field lines between the plates, which is incorrect because the field should be uniform inside. Since both (a) and (c) correctly represent the electric field, we need to check the direction of the field: (a) shows field lines pointing downward (correct) from + + to − −. (c) shows field lines pointing upward (incorrect). ✅ Final Answer: (a) – This correctly represents the electric field between parallel plates.

Question 80

Answer 80

Solution: Finding the Net Electric Field Strength The electric field ( E E) at a point is related to the force experienced by a charge as: E = F q E= q F ​ where: F net F net ​ = net force on the charge, q = 2.0 × 10 − 6 C q=2.0×10 −6 C (charge experiencing the forces), Given forces: F 1 = 3.0 F 1 ​ =3.0 N (leftward) and F 2 = 8.0 F 2 ​ =8.0 N (rightward). Step 1: Find Net Force Since the forces act in opposite directions, we subtract them to find the net force direction: F net = 8.0  N (right) − 3.0  N (left) F net ​ =8.0 N (right)−3.0 N (left) F net = 5.0  N (right) F net ​ =5.0 N (right) Step 2: Find the Electric Field Using: E = F net q E= q F net ​ ​ E = 5.0 2.0 × 10 − 6 E= 2.0×10 −6 5.0 ​ E = 2.5 × 10 6  N/C E=2.5×10 6  N/C Final Answer ✅ (a) 2.5 × 10 6 2.5×10 6 N/C.

Question 81

Answer 81

Solution: Work Done in Moving the Charge The work done ( W W) in moving a charge Q 1 Q 1 ​ from one point to another in an electric field is given by the change in electric potential energy: W = q Δ V W=qΔV where: q = − 2.0 × 10 − 6 C q=−2.0×10 −6 C (charge being moved), Δ V = V T − V S ΔV=V T ​ −V S ​ , where V T V T ​ and V S V S ​ are the potentials at points T and S, respectively. The electric potential due to a point charge is given by: V = k Q r V= r kQ ​ where: k = 9.0 × 10 9 k=9.0×10 9 Nm²/C² (Coulomb’s constant), Q 2 = 8.0 × 10 − 6 C Q 2 ​ =8.0×10 −6 C (source charge), r r = distance from Q 2 Q 2 ​ . Step 1: Compute V S V S ​ and V T V T ​ Potential at Point S ( r S = 5.0 r S ​ =5.0 m): V S = ( 9.0 × 10 9 ) ( 8.0 × 10 − 6 ) 5.0 V S ​ = 5.0 (9.0×10 9 )(8.0×10 −6 ) ​ V S = 72.0 × 10 3 5 V S ​ = 5 72.0×10 3 ​ V S = 1.44 × 10 4 V V S ​ =1.44×10 4 V Potential at Point T ( r T = 7.0 r T ​ =7.0 m): V T = ( 9.0 × 10 9 ) ( 8.0 × 10 − 6 ) 7.0 V T ​ = 7.0 (9.0×10 9 )(8.0×10 −6 ) ​ V T = 72.0 × 10 3 7 V T ​ = 7 72.0×10 3 ​ V T = 1.03 × 10 4 V V T ​ =1.03×10 4 V Step 2: Compute Δ V ΔV Δ V = V T − V S ΔV=V T ​ −V S ​ Δ V = ( 1.03 × 10 4 ) − ( 1.44 × 10 4 ) ΔV=(1.03×10 4 )−(1.44×10 4 ) Δ V = − 4.1 × 10 3 V ΔV=−4.1×10 3 V Step 3: Compute Work Done W = q Δ V W=qΔV W = ( − 2.0 × 10 − 6 ) × ( − 4.1 × 10 3 ) W=(−2.0×10 −6 )×(−4.1×10 3 ) W = 8.2 × 10 − 3 J W=8.2×10 −3 J Final Answer ✅ (b) 8.2 × 10 − 3 8.2×10 −3 J.

Question 82

A 1.60 x 10 -19 C ion is accelerated from rest through a potential difference of 750 V reaching a maximum speed of 8.50 x104 m s. What is the mass of this ion? a. 9.11x 10-31 kg b. 1.67 x 10-27 kg c. 3.32 x 10-26 kg d. 4.84 x 10-20 kg

Answer 82

Solution: Finding the Mass of the Ion The work-energy principle states that the work done by the electric field is converted into kinetic energy: q V = 1 2 m v 2 qV= 2 1 ​ mv 2 where: q = 1.60 × 10 − 19 C q=1.60×10 −19 C (charge of the ion), V = 750 V V=750V (potential difference), v = 8.50 × 10 4 v=8.50×10 4 m/s (final velocity), m m = mass of the ion (to be determined). Step 1: Solve for m m Rearrange the equation: m = 2 q V v 2 m= v 2 2qV ​ Substituting values: m = 2 ( 1.60 × 10 − 19 ) ( 750 ) ( 8.50 × 10 4 ) 2 m= (8.50×10 4 ) 2 2(1.60×10 −19 )(750) ​ m = 2.4 × 10 − 16 7.225 × 10 9 m= 7.225×10 9 2.4×10 −16 ​ m = 3.32 × 10 − 26  kg m=3.32×10 −26  kg Final Answer ✅ (c) 3.32 × 10 − 26 3.32×10 −26 kg.

Question 83

Answer 83

Solution: Identifying the Electric Field Between Two Opposite Charges of Unequal Magnitude Step 1: Understanding the Field Pattern The electric field lines always originate from the positive charge and terminate at the negative charge. If the two charges have equal magnitudes, the field is symmetrical. If the charges have unequal magnitudes, more lines emerge from the stronger (larger magnitude) charge, and the pattern becomes asymmetric. Step 2: Analyzing the Given Diagrams (a): Shows asymmetric field lines, with more lines emerging from one charge than the other, indicating opposite but unequal charges. ✅ Correct (b) & (c): Show equipotential lines, not electric field lines. ❌ Incorrect. (d): Shows two positive charges repelling each other. ❌ Incorrect. ✅ Final Answer: (a) – This correctly represents the electric field between two opposite charges of unequal magnitude.

Question 84

Answer 84

Solution: Finding the Magnitude of the Electric Field at Point P We need to determine the net electric field at point P due to two charges: Q 1 = 8.0 × 10 − 6 C Q 1 ​ =8.0×10 −6 C (positive charge) at x = 0 x=0 Q 2 = − 6.0 × 10 − 6 C Q 2 ​ =−6.0×10 −6 C (negative charge) at x = 2.0 x=2.0 m Point P is 2.0 m to the right of Q 2 Q 2 ​ (so x = 4.0 x=4.0 m) The electric field due to a point charge is: E = k ∣ Q ∣ r 2 E= r 2 k∣Q∣ ​ where: k = 9.0 × 10 9 k=9.0×10 9 Nm²/C² (Coulomb’s constant), r r is the distance from the charge to P. Step 1: Calculate E 1 E 1 ​ (Electric Field Due to Q 1 Q 1 ​ ) E 1 = ( 9.0 × 10 9 ) ( 8.0 × 10 − 6 ) ( 4.0 ) 2 E 1 ​ = (4.0) 2 (9.0×10 9 )(8.0×10 −6 ) ​ E 1 = 72.0 × 10 3 16 E 1 ​ = 16 72.0×10 3 ​ E 1 = 4.5 × 10 3  N/C E 1 ​ =4.5×10 3  N/C Direction: Since Q 1 Q 1 ​ is positive, the field at P points away, meaning it points rightward. Step 2: Calculate E 2 E 2 ​ (Electric Field Due to Q 2 Q 2 ​ ) E 2 = ( 9.0 × 10 9 ) ( 6.0 × 10 − 6 ) ( 2.0 ) 2 E 2 ​ = (2.0) 2 (9.0×10 9 )(6.0×10 −6 ) ​ E 2 = 54.0 × 10 3 4 E 2 ​ = 4 54.0×10 3 ​ E 2 = 1.35 × 10 4  N/C E 2 ​ =1.35×10 4  N/C Direction: Since Q 2 Q 2 ​ is negative, the field at P points toward it, meaning it also points rightward. Step 3: Find the Net Electric Field Since both fields point rightward, we simply add them: E net = E 1 + E 2 E net ​ =E 1 ​ +E 2 ​ E net = ( 4.5 × 10 3 ) + ( 1.35 × 10 4 ) E net ​ =(4.5×10 3 )+(1.35×10 4 ) E net = 1.8 × 10 4  N/C E net ​ =1.8×10 4  N/C Final Answer ✅ (d) 1.8 × 10 4 1.8×10 4 N/C.

Question 85

Answer 85

Solution: Finding the Acceleration of a Proton in an Electric Field The force on a proton due to an electric field is given by: F = q E F=qE where: q = 1.6 × 10 − 19 C q=1.6×10 −19 C (charge of a proton), E = 2.5 × 10 5 E=2.5×10 5 N/C (electric field strength). Using Newton’s Second Law, we find the acceleration: F = m a F=ma a = F m = q E m a= m F ​ = m qE ​ where: m = 1.67 × 10 − 27 m=1.67×10 −27 kg (mass of a proton). Step 1: Compute the Force on the Proton F = ( 1.6 × 10 − 19 ) ( 2.5 × 10 5 ) F=(1.6×10 −19 )(2.5×10 5 ) F = 4.0 × 10 − 14  N F=4.0×10 −14  N Step 2: Compute the Acceleration a = 4.0 × 10 − 14 1.67 × 10 − 27 a= 1.67×10 −27 4.0×10 −14 ​ a = 2.4 × 10 13  m/s 2 a=2.4×10 13  m/s 2 Step 3: Determine the Direction of Acceleration The electric field direction is to the right. Since the proton is positively charged, it experiences a force in the same direction as the field (rightward). Final Answer ✅ (a) 2.4 × 10 13 2.4×10 13 m/s², Right.

Question 86

Answer 86

Solution: Work Needed to Move the Charge The work done to move a charge in an electric field is given by: W = q Δ V W=qΔV where: q = − 2.0 × 10 − 6 C q=−2.0×10 −6 C (charge being moved), Δ V = V T − V S ΔV=V T ​ −V S ​ , where V T V T ​ and V S V S ​ are the potentials at positions T and S, The electric potential due to a point charge is: V = k Q r V= r kQ ​ where: k = 9.0 × 10 9 k=9.0×10 9 Nm²/C² (Coulomb’s constant), Q = 8.0 × 10 − 6 C Q=8.0×10 −6 C (fixed positive charge), r r = distance from the fixed charge. Step 1: Compute V S V S ​ and V T V T ​ Potential at Position S ( r S = 2.0 r S ​ =2.0 m): V S = ( 9.0 × 10 9 ) ( 8.0 × 10 − 6 ) 2.0 V S ​ = 2.0 (9.0×10 9 )(8.0×10 −6 ) ​ V S = 72.0 × 10 3 2 V S ​ = 2 72.0×10 3 ​ V S = 3.6 × 10 4 V V S ​ =3.6×10 4 V Potential at Position T ( r T = 5.0 r T ​ =5.0 m): V T = ( 9.0 × 10 9 ) ( 8.0 × 10 − 6 ) 5.0 V T ​ = 5.0 (9.0×10 9 )(8.0×10 −6 ) ​ V T = 72.0 × 10 3 5 V T ​ = 5 72.0×10 3 ​ V T = 1.44 × 10 4 V V T ​ =1.44×10 4 V Step 2: Compute Δ V ΔV Δ V = V T − V S ΔV=V T ​ −V S ​ Δ V = ( 1.44 × 10 4 ) − ( 3.6 × 10 4 ) ΔV=(1.44×10 4 )−(3.6×10 4 ) Δ V = − 2.16 × 10 4 V ΔV=−2.16×10 4 V Step 3: Compute Work Done W = q Δ V W=qΔV W = ( − 2.0 × 10 − 6 ) × ( − 2.16 × 10 4 ) W=(−2.0×10 −6 )×(−2.16×10 4 ) W = 4.32 × 10 − 2 J W=4.32×10 −2 J Final Answer ✅ (a) 4.3 × 10 − 2 4.3×10 −2 J.

Question 87

Answer 87

Solution: Maximum Kinetic Energy of the Electron The kinetic energy gained by an electron accelerated through a potential difference V V is given by: K E = q V KE=qV where: q = 1.6 × 10 − 19 C q=1.6×10 −19 C (charge of an electron), V = 600 V V=600V (potential difference). Step 1: Compute the Kinetic Energy K E = ( 1.6 × 10 − 19 C ) × ( 600 V ) KE=(1.6×10 −19 C)×(600V) K E = 9.6 × 10 − 17 J KE=9.6×10 −17 J Final Answer ✅ (b) 9.6 × 10 − 17 9.6×10 −17 J.