Module 6: C22 - Electric Fields

Question 1

How can you create an electric field yourself?

Answer 1

You can create an electric field very easily by rubbing a glass rod with a silk cloth. Friction transfers electrons from the glass to the silk, making the cloth negative and the rod positive. The rod is then surrounded by an electric field - it will attract small pieces of paper or water from a dripping tap.

Question 2

How can you test for the presence and strength of an electric field

Answer 2

You can test this using a thin strip of gold foil attached to the bottom of an insulator. The gold foil is given a constant positive charge by momentarily touching it to the charged ball. The charged foil experiences an electrostatic force when close to the ball. This force is smaller the further away the foil is from the charged ball, that is - the field created by the ball is stronger closer to the ball.

Question 3

What is an Electric Field

Answer 3

The area around an object in which a charge would experience a force.

The electric field strength is defined as the force per unit positive charge.

E = F/Q

Question 4

What is the equation for Electric Field Strength

Answer 4

E = F/Q

Electric Field Strength (NC^-1) = Force (N) / Positive Charge (C)

Question 5

Worked Example: Calculating field strength

What is the electric field strength around a point charge if a 3.20 × 10^-19 C charge experiences a force of 7.30 × 10^-15 N?

Answer 5

Q = 3.20x10^-19
F = 7.30x10^-15N

E = F/Q
E = ((7.30x10^-15)/(3.20x10^-19))
E = 2.28x10^4 NC^-1

Question 6

Define electric field strength at a point in space.

Answer 6

It is the force per unit of positive charge.

E = F/Q
(Electric Field = Force / Positive Charge)

Question 7

4 important ideas to know about electric field lines for electric field patterns

Answer 7

● The arrow on an electric field line shows the direction of the field.

● Electric field lines are always at right angles to the surface of a conductor.

● Equally spaced, parallel electric field lines represent a uniform field.

● Closer electric field lines represent greater electric field strength.

Question 8

A proton is accelerated from rest by a uniform electric field of field strength 2.0× 105 NC-1. Assume mass of proton = 1.7 x 10-27 Kg.

Calculate the time it takes to travel a distance of 5.0 cm from its starting point.

Answer 8

E = 2.0x10^5 NC^-1
m = 1.7x10^-27kg

E = F/Q
2x10^5 = F / 1.6x10^-19
F = 3.2x10^-14

F = ma
3.2x10^-14 = 1.7x10^-27a
1.88x10^13 = a

s = 0.05
u = 0
v =
a = 1.88x10^13
t = t

s = ut + 1/2at^2
0.05 = 1/2 x 1.88x10^13 x t^2
0.05 = 9.4x10^12t^2
5.32x10^-15 = t^2
t = 7.29x10^-8 s

Question 9

Is an electric field as scalar or vector quantity

Answer 9

Electric field strength is a vector quantity - it has direction. The direction of the electric field at a point is the direction in which a positive charge would move when placed at that point. Electron fields point away from from positive charges and towards negative charges.

Question 10

Worked Example: Calculating Field Strength

A negatively charged dust particle moves from left to right in a uniform electric field. It has charge of magnitude 5.0x10^-14C and experiences a constant force of 8.6x10^-10N. Calculate the electric field strength and state the direction of the electric field.

Answer 10

Step 1:
The direction of the electric field is the direction in which a positive charge would move - that is, a negative charge will move in the opposite direction. Hence the direction of the electric field is from right to left.

Step 2:
Write down the quantities given in the question.
F = 8.6x10^-10 N,
Q = 5.0x10^-14C,
E = ?

Step 3:
Use the equation for electric field to calculate E.

E = F/Q = 8.6x10^-10 / 5.0x10^-14

E = 1.7x10^4NC^-1

Question 11

What is a Coulomb?

Answer 11

The coulomb (C) is the unit of electric charge.

One coulomb (1C) of charge is defined as the amount of charge that passes a point per second when there is a current of one amp (1A)

Question 12

Coulomb performed an experiment using isolated charges and found that:

‘The force between 2 electric charges Q and q is proportional to Qq and inversely proportional to the square of their separation, r.’

Therefore F α Qq/r2 which implies that F = kQq/r2

Find the units used for the constant of proportionality k

Answer 12

F = k Q/r^2

kgms^-2 = k A^2 S^2 / m^2
kgm^3s^-2 = k A^2 S^2
kgms^3s^-2As^-2 = k
k = kgm^3s^-4A^-2

Question 13

If k has a value of 8.99x109 (units) and is given by the formula k=1/4πε0 find the
value of ε0 known as the permittivity of free space

Answer 13

8.99x10^9 = 1/4πε0
ε0 = 1/4π(8.99x10^9)
ε0 = 8.85x10^-12 FM^-1

Question 14

If k has a value of 8.99x109 (units) and is given by the formula k=1/4πε0.

Calculate the force between an electron and a proton separated by a distance of 5.2 x 10-11m inside an atom.

Answer 14

F = k Qq/r^2
F = 8.99x10^9 x (1.6x10^-19 x 1.6x10^-19) / (5.2x10^11)^2
F = 8.51x10^-8

Question 15

What is the equation for Coulomb’s Law

Answer 15

F = Qq / 4πε0r^2

Question 16

What does Coulomb’s law describe

Answer 16

According to Coulomb’s law, any two point charges exert an electrostatic (electrical) force in each other that is directly proportional to the product of their charge and inversely proportional to the square of the distance between them. This law applies to everything from large charged spheres down to microscopic atoms within all of us.

Question 17

What is ε0 equal to

Answer 17

ε0 = 8.85x10^-12 Fm^-1

Question 18

Calculate the electrostatic force between the following:

An alpha particle (charge = 2e) and a gold nucleus (charge = 79e)
when they are separated by 3.6x10-14 m

Answer 18

F = Qq / 4πε0r^2

F = (2x1.6x10^-19)(79x1.6x10^-19) / 4π x 8.85x10^-12 x (3.6x10^-14)^2 = 28.063N

Question 19

Two spheres with identical charge experience a repulsive force of 2.8µN when they are 12 cm apart. What is the value of the charge?

Answer 19

F = 2.8µN = 2.8x10^-6 N
r = 12cm = 12x10^-2

Q = q

2.8x10^-6 = Q^2 / 4π x 8.85x10^-12 x (12x10^-2)^2
Q = 2.12x10^-9C

Question 20

What is a technique for investigating the forces between two charged metal spheres?

Answer 20

The balls can be charged positively by touching each momentarily to the positive electrode of a high-tension supply. Lowering one of the charged spheres down towards the other will produce a larger reading on the balance.

Question 21

Equation for Electric Field Strength

Answer 21

The electric field strength is defined as the force per unit positive charge

E = F/Q

Question 22

Calculate the electric field intensity at the following places:

0.1 nm from an electron

Answer 22

F = Qq / 4πε0r^2

r = 0.1nm = 1x10^-10m
Q = 1.6x10^-19 C

F = 1.6x10^-19 / 4π x 8.85x10^-12 x (1x10^-10)
F = 1.439x10^11 N

Question 23

What does combing these equations give:

• E = F/Q
• F = Qq/4πε0r^2

Answer 23

E = F/Q => F = EQ
F = Qq/4πε0r^2

EQ = Qq / 4πε0r^2

E = Q / 4πε0r^2

Question 24

Worked Example: Intense Electric Field

The radius of a gold nucleus is about 7.0x10^-15m and it has 79 protons. Estimate the electric field strength at the surface of the nucleus.

Answer 24

Step 1: The charge on a single proton is 1.60x10^-19C. Use this to calculate the charge on the gold nucleus.

Charge Q = 79 x 1.60x10^-19C = 1.264x10^-17C

Step 2: Use the equation for the electric field strength to calculate E.
r = 7.0x10^-15m
Q = 1.264x10^-17C

E = Q / 4πε0r^2
E = 1.264x10^-17 / 4π x 8.85x10^-12 x (7.0x10^-15)^2
E = 2.3x10^21NC^-1

Question 25

For Gravitational Fields and Electric Fields, what is the property that creates the fields

Answer 25

Gravitational Field:
Mass

Electric Field:
Charge

Question 26

For Gravitational Fields and Electric Fields, what type of field is produced

Answer 26

Gravitational Field:
Always attractive (direction of field always towards object)

Electric Field:
Positive point charges produce a repulsive field (direction of field away from object, repels a positive charge) negative point charges produce an attractive field (direction of field towards object, attracts a positive charge).

Question 27

For Gravitational Fields and Electric Fields, what is the field strength (+ equation for it)

Answer 27

Gravitational Field:
Gravitational field strength is force per unit mass
g = F/m

Electric Field:
Electric field strength is the force per unit positive charge
E = F/Q

Question 28

For Gravitational Fields and Electric Fields, what is the force between particles

Answer 28

Gravitational Field:
Force ∝ Product of masses
Force ∝ 1/separation^2

Electric Field:
Force ∝ Product of charge
Force ∝ 1/separation^2

Question 29

For Gravitational Fields and Electric Fields, what are the the force and field strength equations

Answer 29

Gravitational Field:
F = -GMm/r^2
g = - GM/r^2

Electric Field:
F = Qq / 4πε0r^2
E = Q / 4πε0r^2

Question 30

For Gravitational Fields and Electric Fields, what is the type of field

Answer 30

Gravitational Field:
Point masses produce a radial field

Electric Field:
Point charges produce a radial field

Question 31

Is the Electric field between two parallel plate uniform or not?

Answer 31

The electric field between two parallel plates (e.g. capacitor plates) is uniform: a charge will experience the same force wherever it is placed between the plates.

The equation is:
E = V/d

Question 32

Equation for Electric field strength between two capacitor plates (involving voltage and distance)

Answer 32

E = V/d

E = (Vm^-1)

Question 33

How do you derive E = V/d from E = F/Q

Answer 33

E = F/Q (NC^-1)

V = W/Q => Q = W/V
W = Fd => F = W/d

E = (W/d) / (W/V)
E = V/d (Vm^-1)

THEREFORE!
NC^-1 = 1Vm^-1

Question 34

Worked Example:

A Figure shows an arrangement used to accelerate electrons. The separation between the plates is 1.2 cm and the p.d. Across the plates is 3.6kV. Calculate the acceleration of an electron between the plates.

Answer 34

Step 1:
d=1.2x10^-2 m, V= 3.6x103 V

Step 2:
E = V/d = 3.6x10^3/1.2x10-^2 = 3.0x10^5 Vm-1

Step 3:
F = EQ = 3.0x10^5 x 1.60x10^-19= 4.8x10^-14 N

Step 4: F=ma
a = F/m = 4.8x10^-14 / 9.11x10^-31 = 5.3x10^16 ms^-2

Question 35

What does Capacitance on a parallel plate capacitor depend on?

Answer 35

The Capacitance on a parallel plate capacitor depends on the overlap area of the plates, A, and the separation of the plates, d.

Question 36

For plates in a vacuum, what is the relationship between capacitance and area, and capacitance and distance? (+ what equation can you get from it)

Answer 36

Capacitance is proportional to the area (C∝A/d)

Capacitance is inversely proportional to the separation between the plates (C∝1/d)

C ∝ A/d

Question 37

What is the equation for capacitance for a parallel plate capacitor?

Answer 37

C = ε0 / d

Question 38

What is the equation for capacitance for a parallel plate capacitor when a insulator is used rather than a vacuum?

Answer 38

ε = εrε0

Question 39

What is the Equation for Capacitance

Answer 39

C = εoA / d

Question 40

Worked Example:

A capacitor is made from a plastic bin bag sandwiched between two sheets of aluminium foil 60 cm x 30 cm in size. The bin bag is 0.80 mm thick. Plastic has a relative permittivity of 4.0. Calculate the capacitance of this capacitor.

Answer 40

C = ε0εrA/d
C= 8.85 x10^-12x 4.0x(0.60x0.30) / 0.080 x 10^-3 C =7.965x10^-8 F
C = 8.0x10^-8 F

Question 41

In an experiment to measure the charge on an oil drop, the potential difference between two parallel metal plates 5 mm apart was 300 V.

Calculate the electric field strength between the plates.

Answer 41

d = 5mm = 5x10^-3m
V = 300

E = V/d

300 / 5x10^-3 = 60,000NC^-1

Question 42

In an experiment to measure the charge on an oil drop, the potential difference between two parallel metal plates 5 mm apart was 300 V.

Calculate the electrical force on a small oil drop carrying a charge of 3.2x10^-18 C.

Answer 42

E = F/Q
F = EQ

60,000 x 3.2x10^-18 = 1.92x10^-13 N

Question 43

Calculate the energy, in joules, gained by an electron accelerated through a potential difference of 50 kV in an X-ray machine.

Answer 43

50kV = 50,000 V

Q = VQ
W = 50,000 x 1.6x10^-19
W = 8x10^-15 J

Question 44

Calculate the speed of an electron with a kinetic energy of 100 eV.

Answer 44

100eV = 100 x (1.6x10^-19)
= 1.6x10^-17 J

1/2mv^2 = 1.6x10^-17 J
v = 2 x ((1.6x10^-17J) / 9.1x10^-31kg))^1/2
v = 5.93x10^6

Question 45

The capacitance of a parallel plate capacitor is 8.0 pF. State and explain its capacitance when:

a) The separation between the plates is doubled
b) The area of overlap between the plates is halved and their separation is also halved.

Answer 45

C = εA/d

a) The capacitance is halved
b) Capacitance stays the same

Question 46

a) The electric field between the plates just supports the weight of an oil drop of
mass 1.8 × 10-15 kg, which has acquired a charge due to a few excess electrons.
Calculate the charge on the oil drop.

b) The electric field between the plates just supports the weight of an oil drop of
mass 1.8 × 10-15 kg, which has acquired a charge due to a few excess electrons.
Calculate the charge on the oil drop.

Answer 46

a)
Q = mgd / v
Q = 1.8x10^-15 x 9.8 x 3.2x10^-2 / 600
Q = 9.408x10^-19

b)
9.408x10^-19 / 1.6x10^-19 = 5.88 electrons
~ 6 electrons

Question 47

A charge, q, is at rest between two parallel plates.
The plates are then connected to a power supply. Describe the motion of an electron travelling in the direction and opposite to the direction of electric field.

Answer 47

The electron experience a constant electrostatic force because if the uniform electric field between the plates, so it has constant acceleration

Question 48

What 3 ideas/equations can be used to determine the motion of the electron (or any other charged particle) between the plates of a parallel plate capacitor

Answer 48

The following ideas can be used to determine the motion of the electron (or any other charged particle) between the plates:

● Electric field strength E between the plates is given by E=V/d
● Force F in the electron is given by F = EQ = Ee
● Work done on the electron = Vq = Ve

Question 49

Worked Example: Slowing down and stopping

An electron is fired from the positive capacitor plate towards the negative plate along the direction of the electric field with a velocity of 1.0 x 10^7 ms-1. The p.d. Across the plate is 600 V and their separation is 3.0 cm. Calculate the maximum distance the electron will travel.

Answer 49

Step 1: Write down the quantities given in the question.
V = 600, d = 3.0x10^-2m, u = 1.0x10^7ms^-1, v = 0

Step 2: F = Ee = V/d x e
F = 600 x 1.6x10^-19 / 0.030 = 3.2x10^-15N

Step 3: Calculate the magnitude of the declaration of the electron
a = F/m = 3.2x10^-15 / 9.11x10^-31 = 3.513x10^15 ms^-2

Step 4: v^2 = u^2 + 2as, when the electron stops, v=0
s = -u^2/2a = (1.0x10^7)^2 / 2 x (-3.153x10^15) = 1.4x10^-2m

Question 50

Example Question:

Show that a p.d. of 1.5 V can accelerate electrons to 700 kms-1.

Answer 50

Q = 1.6x10^-19
m = 9.11x10^-31

W = VQ
W = 1.5 x 1.6x10^-19
W = 2.4x10^-19 J

W = 1/2mv^2
2.4x10^-19 = 1/2 x 9.11x10^-31v^2
2.4x10^-19 = 4.555x10^-31v^2
5.27x10^11 = v^2
7.25x10^5m = v
V = 725km

V = 700km

Question 51

An electron travelling at 4.6x10^7 ms^-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’ from its path while in-between the two plates?

Answer 51

a)
F = Ve/d
F = 650x1.6x10^-19 / 0.0085

b)
t = d/v
t = 0.036 / 4.6x10^7
t = 7.83x10^-10s

c)
a = EQ/m
F = ma
1.22x10^-14 = 9.11x10^-31a
a = 1.34x10^16

s=s
u=0
v
a=1.34x10^16
t=7.83x10^-10

s = ut+1/2at^2
s = 1/2 x 1.34x10^16 x (7.83x10^-10)^2
s = 4.11x10^-3m

Question 52

What happens to a charged particle if it is moving at a right angle to an electric field?

Answer 52

When an object is projected at a right angle to the electric field, it’s vertical motion is affected, but it’s horizontal motion isn’t.

You can predict the motion of a particle in an electric field by analysing its vertical and horizontal components of motion independently.

Question 53

What applies to a charged particle’s horizontal motion moving in an electric field

Answer 53

• There is no acceleration, hence the horizontal velocity vH of the particle remains constant, with velocity v.
• The time spent in the field is given by the equation t = L/v

Question 54

What applies to a charged particle’s vertical motion moving in an electric field

Answer 54

• The vertical acceleration of the particle is given by the equation: a = F/m = EQ/m
• The initial vertical velocity is u = 0
• The final vertical component of the velocity, Vv as the particle exits the field is give by:
  Vv = u+at = 0 + EQ/m x L/v = EQL/mv

Question 55

What is the area under a force-separation graph

Answer 55

Area under a force - separation graph is the work done.

Question 56

What is the equation for total work done or potential energy for pushing/separating two things

Answer 56

E = Qq / 4πε0r

Question 57

What is electric potential the same as?

Answer 57

The electric potential at a point is defined as the work done per unit charge bringing a positive charge from infinity to that point.

Question 58

Worked Example: Ionic Solid

Common salt is made of sodium, Na+, and chloride Cl-, ions. The separation between each pair of ions is 9.4 x 10-10 m. The magnitude of the charge on each ions is e, 1.6 x 10-9 C.

Estimate the energy in electronvolts (eV) required to pull a pair of ions apart completely. You may ignore the effect of the other ions.

Answer 58

Step 1: Write down the quantities given in the question.
Q = +1.6x10^-19C (sodium)
q = -1.60x10^-19C (chlorine)
r = 9.4x10^-10

Step 2: Use the equations for electric potential energy to calculate E.
E = Qq / 4πε0r
= (1.60x10^-19)(-1.60x10^-19) / 4π x 8.85x10^-12 x 9.4x10^-10
= -2.44x10^-19 J
(The negative sign means that external energy is required to pull these ions apart)

Step 3: Calculate the energy in electron volts (remember 1eV = 1.6x10^-19J)
Energy required = -2.44x10^-19 / 1.60x10^-19 = - 1.5eV

Question 59

Example Question:

A hydrogen atom consists of a proton and an electron orbiting at a distance of 5.29 x 10^-12 m. What is the electrical potential energy (in eV) in this system?

Answer 59

E = Qq / 4πε0r

E = (1.6x10^-19)(-1.6x10^-19) / 4π x 8.85x10^-12 x 5.29x10^-12

E = -4.35x10^-17 J

4.35x10^-17 / 1.6x10^-19 = -272eV

V = -272eV

Question 60

Example Question:

What is the electric potential at a distance of 20.0 mm around a point charge
of 1.40 × 10^-8 C?

Answer 60

V = Q / 4πε0r

V = 1.40x10^8 / 4π x 8.85x10^-12 x 0.02

V = 6294V

Question 61

(a) What is the electrical potential in the field of a +5nC point charge at a distance of 0.75m from the charge?

b) A charge of +2nC is placed into the field at this point. What is the potential energy of the system?

c) If the +2nC charge is now moved through the field to a point 0.25m from the first charge, how much work has been done by the force moving the charge?

Answer 61

a)
V = Q / 4πε0r
V = 5x10^-9 / 4π x 8.85x10^-12 x 0.75
V = 59.95V

b)
E = Vq
(2x10^-9)(59.9) = 1.20x10^-7 J

c)
E = Qq / 4πε0r
E = (5x10^-9)(2x10^-9) / 4π x 8.85x10^-12 x 0.25
E = 3.6x10^-7
3.6x10^-7 - 1.20x10^-7 = 2.4x10^-7

Question 62

Equations for Capacitance of a Charged Sphere

Answer 62

C = Q/V
= Q / Q / 4πε0r
= 4πε0r

Question 63

What are Equipotential Lines?

Answer 63

Equipotential lines link points of equal electric potential in an electric field.