Electric Potential Energy

Electric Potential Energy of Two Point Charges

Work Done

Work is done when:
- A positive charge moves in the same direction as the electric field lines in an electric field
- A negative charge moves in the opposite direction to the electric field lines in an electric field
The work done in moving a charge Q is equal to the electric potential energy
The work done is calculated by the equation:

$W = Q V$

Where:
- W = work done (J)
- V = electric potential due to a point charge (V)
- Q = charge (C)

The electric potential energy E_p at a point in an electric field is defined as:

The work done in bringing a charge from infinity to that point

The electric potential energy of a pair of point charges Q₁(charge that is producing the electric field) and Q₂ (charge that is moving in the electric field) is defined by:

$E_{p} = \frac{Q_{1} Q_{2}}{4 π ε_{0} r}$

Where:
- E_p = electric potential energy (J)
- Q₁ = charge that is producing the electric field (C)
- Q₂ = charge that is moving in the electric field (C)
- r = separation of the charges Q₁ and Q₂ (m)
- ε₀ = permittivity of free space (F m^-1)
Unlike the electric potential, the electric potential energy will always be positive
Recall that at infinity, V = 0 therefore E_p = 0

Electric Potential Energy for a Moving Positive and Negative Charge

Change in Electric Potential Energy, downloadable AS & A Level Physics revision notes

Work is done when moving a point charge away from another charge. E.P.E is another way of saying electric potential energy

Change in Electric Potential Energy

It is more useful to find the change in electric potential energy when one charge moves away from another
The change in electric potential energy from a charge Q₁ at a distance r₁ from the centre of charge Q₂ to a distance r₂ is given by the equation:

$∆ E_{p} = \frac{Q_{1} Q_{2}}{4 π ε_{0}} (\frac{1}{r_{1}} - \frac{1}{r_{2}})$

The change in electric potential, ΔV, from charge Q₁ is:

$∆ V = \frac{Q_{1}}{4 π ε_{0}} (\frac{1}{r_{1}} - \frac{1}{r_{2}})$

Both equations are similar to the change in gravitational potential between two points near a point mass

Worked example

An α-particle ${}_{2}^{4}{He}$ is moving directly towards a stationary gold nucleus ${}_{79}^{197}{Au}$ .

At a distance of 4.7 × 10⁻¹⁵ m the α-particle momentarily comes to rest.

Calculate the electric potential energy of the particles at this instant.

Answer:

Step 1: Write down the known quantities

Distance, r = 4.7 × 10^-15 m
The charge of one proton = +1.60 × 10^-19 C
An alpha particle (helium nucleus) has 2 protons
- So, charge of alpha particle, Q₁ = 2 × 1.60 × 10^-19 = +3.2 × 10^-19 C
The gold nucleus has 79 protons
- So, charge of gold nucleus, Q₂ = 79 × 1.60 × 10^-19 = +1.264 × 10^-17 C

Step 2: Write down the equation for electric potential energy

$E_{p} = \frac{Q_{1} Q_{2}}{4 π ε_{0} r}$

Step 3: Substitute values into the equation

$E_{p} = \frac{(1.264 \times 10^{- 17}) \times (3.2 \times 10^{- 19})}{4 π \times (8.85 \times 10^{- 12}) \times (4.7 \times 10^{- 15})} = 7.7 \times 10^{- 12} J (2 s . f .)$

Exam Tip

This topic has a lot of confusing concepts and language. Make time to take notes on this topic and learn the correct equations for each quantity. When calculating electric potential energy, make sure you do not square the distance!

Electric Potential Energy (CIE A Level Physics)

Revision Note