Equipotential surfaces

Walking along a contour

A hiking map has contour lines. Each line joins places of the same height. If you walk along one, you never climb, and gravity does no work on you. If you walk straight across the lines, you climb or descend, and the lines are close together where the slope is steep.

The electric field has contours too. They are called equipotential surfaces. A surface is an equipotential surface if the potential has the same value at every point of it.

No work along the surface

Move a charge $q$ between two points A and B of the same equipotential surface. The work is $W = q(V_B - V_A)$, and $V_B - V_A = 0$. So no work is done in moving a charge along an equipotential surface. It does not matter which path we choose between the points. Only the potentials at the end points count.

The field is perpendicular to the surface

Suppose the field made an angle with the surface that was not $90^\circ$. Then the field would have a part along the surface. That part would push on a charge as it slides along the surface, and it would do work. But we have just shown that the work is zero. So the part along the surface must be zero:

At every point, the electric field is perpendicular to the equipotential surface through that point.

Two more facts follow.

• The field points from higher potential to lower potential. If you go along a field line, the potential falls.
• Two equipotential surfaces never cross. At a crossing point the potential would need two different values, and the field would need two different directions.

The shapes of equipotential surfaces

• A point charge. $V = kq/r$ has the same value at every point of the same $r$. The surfaces are concentric spheres, and the field lines are straight lines along the radius.
• A uniform field. $V$ changes only along the field. The surfaces are flat planes, perpendicular to the field, at equal gaps for equal steps of $V$.
• A dipole. The surfaces are curved round each charge. The plane halfway between the charges, which is perpendicular to the axis, is at $V = 0$.

Check that every purple field line crosses every dashed green line at a right angle.

Spacing shows the strength

Draw the surfaces at equal steps of potential, such as every $10$ V. Around a point charge they crowd together near the charge and spread out far away. Between plates they stay evenly spaced.

Where the surfaces are close, the potential changes quickly over a short distance. There the field is strong. Where they are far apart, the field is weak. A map of equipotential surfaces shows the strength of the field as well as its direction.

The field from the potential

Move a small step $\Delta l$ along a field line, in the direction of the field. The field does the work $qE\,\Delta l$ on a charge $q$. The potential energy falls by this amount, so $q\,\Delta V = -qE\,\Delta l$. Cancel $q$, and make the step very small:

The minus sign says that the field points in the direction in which the potential falls. For a step in any direction, the component of the field in that direction is $E_l = -\dfrac{dV}{dl}$. Along the $x$ direction, $E_x = -\dfrac{dV}{dx}$.

Check it on a point charge. $V = kq/r$ gives $E = -\dfrac{d}{dr}\left(\dfrac{kq}{r}\right) = \dfrac{kq}{r^2}$. This is Coulomb's law again.

Check it between plates. Take $x$ along the line from the negative plate to the positive plate, with $V = (3000\ \text{V/m})\,x$. Then $E_x = -3000$ V/m. The minus sign says that the field points towards the negative plate, which is right.

A worked example. Along a line, the potential is $V = 5.0x^2 - 8.0x$, in volt, with $x$ in metres.

1. The field. $E_x = -\dfrac{dV}{dx} = -(10x - 8.0)\ \text{V/m} = (8.0 - 10x)\ \text{V/m}$.
2. At $x = 0.30$ m. $E_x = 8.0 - 3.0 = 5.0\ \text{V/m}$. It points in the $+x$ direction.
3. Where is the field zero? $8.0 - 10x = 0$ gives $x = 0.80$ m. The potential there is $5.0 \times 0.64 - 8.0 \times 0.80 = -3.2$ V, the lowest value of $V$ along the line. The field is zero where the graph of $V$ is flat.

If the potential depends on three coordinates, each component of the field is found in the same way, from the rate of change along its own axis. The other coordinates are held fixed:

For example, let $V = 4x^2 - 3xy + 2y$ in volt, with $x$ and $y$ in metres. At $x = 1.0$ m and $y = 2.0$ m, $E_x = -(8x - 3y) = -(8.0 - 6.0) = -2.0$ V/m. Also $E_y = -(-3x + 2) = 1.0$ V/m. The size of the field is $\sqrt{(-2.0)^2 + (1.0)^2}\ \text{V/m} = 2.2\ \text{V/m}$.

Work between different surfaces

If A and B are on two different surfaces, the work to carry a charge from A to B is $q(V_B - V_A)$. It does not depend on the path. A long wandering path gives the same work as a straight one. All that matters is the potential difference between the two surfaces.

In a metal that carries no current, every point has the same potential, so the surface of a conductor is an equipotential surface. We shall see why when we study conductors.

Choose a larger charge, and look at the field at the same distance. Then move the point out. Does the gap between neighbouring surfaces get wider or narrower?

Practise with new numbers

Try it in Python

The program draws the equipotential lines of a $+4$ nC and a $-4$ nC charge. It calculates the potential at each point of a grid, and a contour plot joins the points that have the same value.

Common mistakes

• Thinking the field is along the surface. It is perpendicular to it.
• Thinking that no work means no force. On an equipotential surface the force is not zero. It is the work that is zero, because the force is perpendicular to the motion.
• Forgetting the minus sign in $E = -dV/dx$. It says that the field points to lower potential.
• Using the gap in cm. For $E = \Delta V/\Delta l$ in V/m, change the gap to m.
• Drawing two equipotential surfaces that cross.
• Saying equal spacing is always true. Equal steps of potential are evenly spaced only in a uniform field.

Remember these

• An equipotential surface has the same $V$ everywhere on it. 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 has $V = 0$.
• Close surfaces mean a strong field: $E = \dfrac{\Delta V}{\Delta l}$.
• $E_l = -\dfrac{dV}{dl}$. For three coordinates, $E_x = -\dfrac{\partial V}{\partial x}$, and likewise for $y$ and $z$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What shape are the equipotential surfaces of a single point charge, and of a uniform field?

2. Two equipotential lines, $2.0$ V apart, are $0.50$ cm apart. What is the size of the field between them?

3. A charge is carried along an equipotential surface, in a region where the field is strong. How much work does the field do?

4. Along a line, the potential falls steadily from $90$ V at $x = 0.10$ m to $60$ V at $x = 0.40$ m. What are the size and the direction of the field?

5. Why can two equipotential surfaces never cross?