D2 - Electric and Magnetic Fields

Question 1

what are the steps to set up an electric fields demo

Answer 1

1. set up a high potential difference across 2 plates
2. sprinkle dipolar grains (semolina) on castor oil (this allows the grains to move)
3. observe the movement of the grains

Question 2

what are the 2 types of electric fields

Answer 2

radial
uniform plates

Question 3

what is the strength for a uniform plate

Answer 3

the field has a constant field strength everywhere, because the distance between the plates are always the same

Question 4

which way do the lines go from in uniform plates

Answer 4

from positive to negative

Question 5

what are the 3 things you need to draw to get the three marks for a uniform plate drawing

Answer 5

1. the field line between the plates should be equally spaced
2. there should be arrows on the field lines going from positive to negative with the labelled plates
3. you should have more than 3 field lines drawn between the plates

Question 6

how do you draw the field lines for a positive radial charge

Answer 6

1. draw a positively charged particle in the centre
2. draw about 12 lines coming out of the particle at equal intervals (it should look a bit like a spider diagram)
3. add the arrows going out of the particle

Question 7

where is the force strongest on a radial charge

Answer 7

it is strongest closer to the particle

Question 7

how do you draw the field lines for a negative radial charge

Answer 7

1. draw a negatively charged particle in the centre
2. draw about 12 lines coming out of the particle at equal intervals (it should look a bit like a spider diagram)
3. add the arrows going in of the particle

Question 8

where is the force weakest on a radial charge

Answer 8

it is weakest further away from the particle

Question 9

what is the definition of electric field strength

Answer 9

the force per unit charge on a small positive test charge

Question 10

can a positive charge survive without a negative charge in an electric field? how about for a magnetic field

Answer 10

electric field - positive can live without negative charge
magnetic - need both positive and negative poles to create a magnetic field

Question 11

(EXAM Q)
Electric fields are caused by both point charges and by parallel plates with a pd across them. Describe the difference between the electric field caused by a point charge and the electric field between the parallel plates. your answer should include a diagram of each type of field and a reference to electric field strength

Answer 11

An electric field caused by a point charge has a stronger electric field strength closer to the point charge, that becomes weaker further away from charge because it is more spread out. The electric field caused by parallel plates is uniform, as the distance between the plates is constant. Therefore the electric field is not stronger or weaker at any specific point

Diagram drawing:
Parallel plates - draw 4 lines between two thin rectangles (the plates). Draw a positive sign on one plate and a negative on the other. Draw arrows going from positive to negative on the field lines

Point charge - draw a positive charge with 8-10 lines coming out of it perpendicular to the surface of the charge. Draw arrows coming out of the charge on the field lines

Question 12

Whats the difference between a field force and a contact force

Answer 12

contact - physical contact
field - an object only experiences this when in the charges field

Question 13

what does the electric field for an electron look like

Answer 13

A negative charge with field lines going into in. With arrows on the lines going into the charge

Question 14

which charge do we use when calculating electric field strength

Answer 14

the charge affected by the force

Question 15

What does an electric field between 2 parallel plates depend on

Answer 15

1. the potential difference between the plates (V)
2. the distance between them (d)

Question 16

what is the formula for electric field strength between parallel plates

Answer 16

E = V/d

Question 17

what is the charge of an electron? proton?

Answer 17

-1.6x10^-19C
1.6x10^-19C

Question 18

what is the formula for potential energy for an electron

Answer 18

Ep = VQ

Question 19

A charged particle of 8x10^-19C is held stationary inside an electric field produced by two electric plates. The voltage between the plates is 270V and they are separated by a distance of 6x10^-3m.

a) what constant field strength exists between the plates?
b) What is the mass of the particle? (Hint: first draw a free-body diagram of the particle to determine its weight)

Answer 19

a) E = V/d
E = 270/ 6x10^-3
E = 45000NC^-1

b) ‘stationary’ means the Weight equals the electric force.
Electric force = EQ
mg = EQ
m = EQ/g
m = (45000)(8x10^-19)/9.81
m = 3.7x10^-15 kg

Question 20

A proton travelling at 3.4 x 10^6 ms^-1 passes through an electric field. How fast will the proton be going after it emerges from the field?

Answer 20

Ep = VQ
Ep = 1.2 x 10^-15J
Ke = 1/2mv^2
Ke = 9.7 x 10^-15J
Ep + Ke = 1.09 x 10^-14J
v = square root (2VQ/m)
because Ep = VQ (total energy),
v = square root (2Ep/m)
v = square root (2)(1.09 x 10^-14)/1.673 x 10 ^-27
v = 3.6 x 10^6 ms^-1

Question 21

An electron travelling at 4.6 x 10^7ms^-1 enters a constant electric field between two charged plates spread 0.0085m apart. The voltage between the plates is 650V and the plates are 0.036m long.

a) what is the electric force acting on the electron
b) how much time is taken for the electron to pass through the plates
c) how far will the electron fall in its path, while in between the two plates

Answer 21

a) V/d = 76470.588 = E
F = E x q = 1.22 x 10 ^-4N

b) time = distance/ speed
time = 0.036/ 4.6 x 10^7
time = 7.83 x 10^-10

c)
s ?
u 4.6 x 10^7
v X
a F/m = 1.22 x10^-14/ 9.1 x 10^-31 = 1.34 x 10^16
t 7.83 x 10^-10

s = ut +1/2at^2
s = 0.04m

Question 22

what is coulombs law

Answer 22

The electrical force between two stationary point charges where the force is inversely proportional to the square of the separation

Question 23

what is a point charge

Answer 23

a particle of zero size that carries an electric charge

Question 24

what is coulombs law equation

Answer 24

F = k(q1q2)/r^2

Question 25

the force on q1 is …… in direction to the force on q2

Answer 25

equal in magnitude and opposite in direction

The force of 2 on 1 is equal but opposite to the force of 1 on 2

Question 26

what is electric potential energy + formula

Answer 26

the potential energy a charged particle has, due to the force it experiences from the field

Ep = kQq/r

Question 27

how do we find the work done to separate two attracting charges?

Answer 27

you need to find the electric potential energy at the start point and at the end point and minus them

Question 28

Find the work done to move a charge (q) from 3m to 0.5m under the influence of charge (Q)

charge q = 3 x 10^-6
charge Q = 20 x 10^-6

Answer 28

Ep 1 = k(20x10^-6)(3x10^-6)/3
Ep 2 = k(20x10^-6)(3x10^-6)/0.5

Ep1 - Ep2 = 0.899J = 0.9J

Question 29

What is electric potential (V) + formula

Answer 29

the electrical potential energy per unit charge at a point in a field

V = Wd/q = kQ/r

Question 30

what is the electric potential at a point 2m from a +3C charge

Answer 30

V = kQ/r
V = (8.99x10^9)(3)/2
V = 1.35x10^10JC^-1

Question 31

what is the potential halfway between a +3C charge and a -3C charge separated by 1m

Answer 31

No potential

Question 32

Find the electric potential at point B due to the other charges:

A = +30x10^-6
C = +70x10^-6
distance between A and B = 20cm
distance between B and C = 60cm

Answer 32

V1 = kq/r (A)
V1 = 1348500

V2 = kq/r (A)
V2 = 1048833.333

V1 + V2 = 2.4x10^6V

Question 33

how much energy does it take to move along an equipotential line

Answer 33

no energy

Question 34

what is the gradient of the graph of electric potential

Answer 34

electric field strength