Potential & Potential Energy

Work Done on a Mass

When a mass is moved against the force of gravity, work is done
The work done in moving a mass m is given by:

Δ W = m \times Δ V

Where:

∆W = change in work done (J)
m = mass (kg)
∆V = change in gravitational potential (J kg^-1)

This change in work done is equal to the change in gravitational potential energy (G.P.E)

When V = 0, then the G.P.E = 0

The change in G.P.E, or work done, for an object of mass m at a distance r₁ from the centre of a larger mass M, to a distance of r₂ further away can be written as:

Δ G . P . E = - \frac{G M m}{r_{2}} - (- \frac{G M m}{r_{1}}) = G M m (\frac{1}{r_{1}} - \frac{1}{r_{2}})

- Where:
  - M = mass that is producing the gravitational field (eg. a planet) (kg)
  - m = mass that is moving in the gravitational field (eg. a satellite) (kg)
  - r₁ = first distance of m from the centre of M (m)
  - r₂ = second distance of m from the centre of M (m)

Work is done when an object in a planet's gravitational field moves against the gravitational field lines i.e.: away from the planet

Change in GPE

Gravitational potential energy increases as a satellite leaves the surface of the Moon

Worked example

A spacecraft of mass 300 kg leaves the surface of Mars to an altitude of 700 km. Calculate the work done by the spacecraft. The radius of Mars = 3400 km, Mass of Mars = 6.40 × 10²³kg

Step 1: Write down the work done (or change in G.P.E) equation

Δ G . P . E = G M m (\frac{1}{r_{1}} - \frac{1}{r_{2}})

Step 2: Determine values for r₁ and r₂

r₁ is the radius of Mars = 3400 km = 3400 × 10³ m

r₂ is the radius + altitude = 3400 + 700 = 4100 km = 4100 × 10³ m

Step 3: Substitute in values

Δ G . P . E = (6.67 \times 10^{- 11}) \times (6.40 \times 10^{23}) \times 300 \times (\frac{1}{3400 \times 10^{3}} - \frac{1}{4100 \times 10^{3}})

Δ G . P . E = 643.076 \times 10^{6} = 640 M J

Work Done on a Charge

When a mass with charge moves through an electric field, work is done
The work done in moving a charge q is given by:

Δ W = q \times Δ V

Where:

∆W = change in work done (J)
q = charge (C)
∆V = change in electric potential (J C^-1)

This change in work done is equal to the change in electric potential energy (E.P.E)

When V = 0, then the E.P.E = 0

The change in E.P.E, or work done, for a point charge q at a distance r₁ from the centre of a larger charge Q, to a distance of r₂ further away can be written as:

Δ E . P . E = \frac{Q q}{4 π ε_{o}} (\frac{1}{r_{2}} - \frac{1}{r_{1}})

- Where:
  - Q = charge that is producing the electric field (C)
  - q = charge that is moving in the electric field (C)
  - r₁ = first distance of q from the centre of Q (m)
  - r₂ = second distance of q from the centre of Q (m)

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

Work is done when moving a point charge away from another charge

Work is done when a positive charge in an electric field moves against the electric field lines or when a negative charge moves with the electric field lines

Worked example

The potentials at points R and S due to the +7.0 nC charge are 675 V and 850 V respectively.

Work Done Electric Field Worked Example, downloadable AS & A Level Physics revision notes

Calculate how much work is done when a +3.0 nC charge is moved from R to S.

Step 1: Write down the known quantities

- p.d. at R, V₁ = 675 V
- p.d. at S, V₂ = 850 V
- Charge, q = +3.0 nC = +3.0 × 10^-9 C

Step 2: Write down the work done equation

W = q × ΔV

Step 3: Substitute in the values into the equation

W = (3.0 × 10^-9) × (850 - 675) = 5.3 × 10^-7 J

Examiner Tip

Make sure to not confuse the ΔG.P.E equation with ΔG.P.E = mgΔh, they look similar but refer to quite different situations.

The more familiar equation is only relevant for an object lifted in a uniform gravitational field, meaning very close to the Earth’s surface, where we can model the field as uniform.

The new equation for G.P.E does not include g. The gravitational field strength, which is different on different planets, does not remain constant as the distance from the surface increases. Gravitational field strength falls away according to the inverse square law.

Remember that q in the work done equation is the charge that is being moved, whilst Q is the charge which is producing the potential. Make sure not to get these two mixed up. It is common for both to be given in the question, as in our worked example. You are expected to choose the correct one.

Potential & Potential Energy (DP IB Physics: HL)

Revision Note