Electric Potential and Equipotentials: Quick Revision

A brief note on Electric Potential and Equipotentials. Read it before a test. Try each question at the end before you open its answer.

Potential energy and potential

• Potential difference: $V_B - V_A = \dfrac{W_{AB}}{q_0}$, the work to carry a unit positive charge slowly from A to B. The unit is the volt, $1\ \text{V} = 1\ \text{J/C}$.
• Potential at a point: the work to bring a unit positive charge from infinity to that point. It is a scalar.
• Potential energy of a charge: $U = qV$. Positive charges fall to lower potential, and negative charges to higher potential.
• Uniform field between plates: $E = \dfrac{V}{d}$, in V/m. The potential falls along the field.
• Electronvolt: $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$. A charge $q$ falling through $V$ gains $\tfrac{1}{2}mv^2 = qV$.

Potential of charges

• Point charge: $V = \dfrac{kq}{r}$, with $k = \dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2$. The sign of $q$ counts.
• Many charges: $V = k\sum \dfrac{q_i}{r_i}$. Add the numbers.
• Ring on its axis: $V = \dfrac{kQ}{\sqrt{R^2 + x^2}}$. Shell: $\dfrac{kQ}{r}$ outside, $\dfrac{kQ}{R}$ inside.
• Dipole: $V = \dfrac{kp\cos\theta}{r^2}$ far away. On the equatorial line $V = 0$.
• Energy of a pair: $U = \dfrac{kq_1q_2}{r}$. A group has the sum over every pair.
• Dipole in a field: $\tau = pE\sin\theta$ and $U = -pE\cos\theta$.

Equipotential surfaces

• The potential is the same everywhere on the surface, so no work is done along it.
• The field is perpendicular to the surface and points to lower potential.
• Point charge: concentric spheres. Uniform field: parallel planes. Dipole: the middle plane is at $0$ V.
• Field from potential: $E_l = -\dfrac{dV}{dl}$. Close surfaces mean a strong field, $E = \dfrac{\Delta V}{\Delta l}$.

A worked example

A point charge of $+2.5$ nC is in air. A charge of $-4.0$ nC is placed $30$ cm from it. Find the potential there and the potential energy of the $-4.0$ nC charge.

1. The potential. $V = \dfrac{kq}{r} = \dfrac{9.0 \times 10^{9} \times 2.5 \times 10^{-9}}{0.30}\ \text{V} = 75\ \text{V}$.
2. The energy. $U = qV = (-4.0 \times 10^{-9}\ \text{C}) \times 75\ \text{V} = -3.0 \times 10^{-7}\ \text{J}$. The negative sign says that the two charges attract.

Practise with new numbers

Check yourself

1. A point charge makes a potential of $80$ V at a distance of $0.60$ m. What is the potential at $1.2$ m?

2. Two equipotential lines that are $1.5$ V apart lie $0.30$ cm apart. What is the size of the field between them?

3. Why is no work done when a charge moves along an equipotential surface, and what does that say about the field?