5.1 Electric Potential Difference

IB Physics Power Points

Topic 5.0

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Electric Potential Energy Difference

Electric Potential Energy Difference

Electric Potential Energy Difference

Recall: an object held above the Earth’s surface has a gravitational potential energy associated with its position.

Similarly: a charge placed in an electric field will have an electrical potential energy associated with its position.

electric potential energy difference ΔPEelectric and electric potential difference ΔV can be difficult concepts to keep straight. As we proceed, make sure you ensure you understand these concepts.

Electrical Potential Energy in a Uniform Electrical FieldThe electric field between two parallel

plates is an example of a uniform electric field.

A positive charge placed at Point P will feel an electrical force directed towards the negative plate.

P

Due to its location in the field the charge has electrical potential energy.

If the charge is released, it will accelerate towards the negative plate. The PEelectric at the positive plate is transferred to kinetic energy (motion)

+

E

Connection : freefall in a gravitational field

The change in potential energy for a positive charge placed at Point P if it moves to the negative plate can be expressed as:

The negative sign indicates that the amount of PE is decreasing. Work is being done on the charge by field.

elecPE Eq x

This equation only applies to a uniform electric field. WHY?

P+

E

Derive this equation!

Electrical Potential Energy in a Uniform Electrical Field

elecPE Eq x

+

E = 2 NC-1

+

-

-

How does the potential energy of each charge change with a change in position as indicated in the diagram?

Calculate the change in PE in each scenario.

Electrical Potential DifferenceElectrical potential difference is defined as the work done per unit of charge (Joules per Coulomb) while moving the charge between two points in an electric field. It can also be thought of as the change in PE per unit of charge between two points in an electric field.

in data booklet it is written as V

Ex

The negative sign indicates that the direction of decreasing potential is the same as the field direction.

The potential difference between the positive and the negative plates depends on the field strength and the plate separation. Potential difference is measured in volts (1 V = 1 JC-1).

Positive p

late

Negative p

late

+ Dx

Highpotential

Lowpotential

E (field strength)

Change in PE = -(work done by field)

electricPE WV E x

q q

- Dx

Consider an electron being placed in an electrical field across which a potential difference of 1 V is measured.P

ositiv

e p

late

Neg

ativ

e p

late

-Δx

Highpotenti

al

Lowpotenti

al

E

-e

191.6 10 Joules electricPE Vq

Potential

difference = 1 V

Note the following:

The potential energy of the electron DECREASES by 1.6 x 10-19 Joules. The kinetic energy would increase by the same amount.

1.6 x 10-19 Joules = 1 electron volt. An eV is a unit of energy equivalent to the energy change of an electron accelerated through a potential difference of one volt.

The eV is used mainly to describe small amounts of energy.

Problems to try:It takes 4.2 x 10-3 J of work to move 1.2 x 10-6 C of charge from point X to point Y in a uniform electric field. What is the potential difference between X and Y. W

Vq

3

6

4.2 10

1.2 10

V

3500 V V

Technically, the answer is +3500 V. If work is done by an external force against the field, then the potential energy of the charge being moved is increasing.

Problems to try:Calculate the magnitude of the electric field in a parallel plate apparatus whose plates are 5.0 mm apart and have a potential difference of 300 V between them.

V

Ex

300

0.0050E

4 -16.0 10 NC E

Problems to try:What potential difference would have to be maintained across the plates of a parallel plate apparatus if they are 1.2 cm apart to create an electric field intensity of 1.5 x 104 NC-1?

V E x

41.5 10 (0.012) V

-1180 JC V

Do dimensional analysis to show that a Volt is indeed equal to one JC-1.

Problems to try:Calculate the work done in moving a 10.0 C charge through a potential difference of 150 V?

W q V

610.0 10 (150) W

0.0015 JW

It does not matter if the work is done by an external force against the field (increase in charge PE) or by the field itself (decrease in charge PE)

Problems to try:How far apart must two parallel charged plates be situated so that a potential difference of 1.50 x 102 V across the plates produces an electric field strength of 1.00 x 103 NC-1?

Vx

E150

1000 x

0.150 m x