6.2.4 - Electric potential, energy and isolated spheres

Potential from infinity

Electric potential is the electric version of "energy per unit charge". It is useful because the field around a source charge may be complicated, but once you know the potential at a point, the energy of a charge placed there is found by multiplying by that charge.

The unit of electric potential is the volt, $\mathrm{V}$. Since it is work done per unit charge,

The phrase "unit positive charge" matters. Potential is defined using a positive test charge, so a positive source charge gives a positive potential and a negative source charge gives a negative potential. Potential is a scalar: it has a sign, but it is not a vector direction.

Infinity is chosen as the zero because, very far from an isolated point charge or spherical charge, the field becomes negligible and the potential tends to zero. This is like choosing a reference level for gravitational potential energy: the value only makes sense once the zero has been chosen.

Do not confuse electric potential with electric field strength. Electric field strength is force per unit charge and is a vector, measured in $\mathrm{N}\,\mathrm{C}^{-1}$ or $\mathrm{V}\,\mathrm{m}^{-1}$. Electric potential is energy per unit charge and is a scalar, measured in $\mathrm{J}\,\mathrm{C}^{-1}$ or $\mathrm{V}$.

Potential near a point charge

For a point charge, or for a charged sphere when you are outside the sphere, the electric potential depends on distance from the centre of the charge distribution.

Here $V$ is the electric potential at the point in volts, $Q$ is the source charge in coulombs, $r$ is the distance from the source charge in metres, and $\varepsilon_0$ is the permittivity of free space. The factor $1/(4\pi\varepsilon_0)$ is the same constant that appears in Coulomb's law.

There are three important consequences.

First, the sign of $V$ follows the sign of $Q$. A positive source charge gives a positive potential; a negative source charge gives a negative potential.

Second, the magnitude of $V$ decreases as $r$ increases. The potential tends towards zero at infinity, so a point close to a positive source charge is at a higher potential than a point further away.

Third, this equation is not the uniform-field equation. The equation $E = V/d$ belongs to a uniform field between parallel plates. Around a point charge or isolated sphere, the field is radial and changes with distance, so you must use the point-charge potential equation for $V$.

For changes in potential between two distances:

Worked example: change in potential

A small object has charge $Q = +8.0\,\mathrm{nC}$. Point A is $0.20\,\mathrm{m}$ from the object and point B is $0.50\,\mathrm{m}$ from it. Determine the change in electric potential from A to B.

Convert the charge:

Use $1/(4\pi\varepsilon_0) = 8.99 \times 10^9\ \text{N m}^2\text{C}^{-2}$.

At A:

At B:

So

The negative sign means the potential has decreased as the point has moved further from the positive charge.

Isolated sphere capacitance

An isolated conducting sphere can store charge. The charge spreads over the outside surface, and outside the sphere the electric field and potential are the same as if all the charge were concentrated at the centre.

The capacitance of any conductor is

where $Q$ is the charge on the conductor and $V$ is its potential relative to infinity.

For a conducting sphere of radius $R$, the surface is at distance $R$ from the centre. Use the point-charge potential equation at the surface:

Now substitute this into $C = Q/V$:

The charge $Q$ cancels:

This result has a simple meaning: a larger isolated sphere has a larger capacitance. It can hold more charge for the same potential because the charge is spread over a larger conductor and the surface is further from the centre.

Be careful with the word "isolated". It means the sphere is far from other conductors and charge distributions, so infinity can act as the zero-potential reference. It is not a parallel-plate capacitor, and it does not have a second nearby plate.

Worked example: OCR-style sphere calculation

A metal sphere has radius $1.5\,\mathrm{cm}$ and is charged to a potential of $+5.0\,\mathrm{kV}$. Calculate the charge on the sphere.

Convert the values:

First calculate the capacitance:

Then use $Q = CV$:

The answer is small because even a conducting sphere has a very small capacitance unless its radius is very large.