Equipotential Surfaces and Field Lines

Equipotential Surfaces and Field Lines — A Complete Guide

In every electrostatics problem, the same two pictures appear: field lines and equipotential surfaces. They’re two ways of seeing the same physics. Master the relationship between them and the rest of the chapter — capacitance, dielectrics, work done — becomes far easier.

The single big idea: field lines are always perpendicular to equipotential surfaces. If they weren’t, work would be done moving a charge along a “constant-potential” surface, which is a contradiction.

Key Terms & Definitions

Equipotential surface — a surface (or curve) on which the electric potential $V$ has the same value at every point. Moving a charge from one point on the surface to another requires zero work.

Electric potential $V$ — the work done per unit positive test charge to bring it from infinity (or a chosen reference) to the point. Units: volts (V).

Electric field line — a directed curve along which a positive test charge would move if released from rest. The tangent to a field line at any point gives the direction of $\vec{E}$ there.

Potential difference $\Delta V = V_B - V_A$ — the work done per unit charge to move a positive test charge from $A$ to $B$ .

Methods & Concepts

The fundamental relationship

For an infinitesimal displacement $d\vec{r}$ along an equipotential surface, $V$ doesn’t change, so $dV = 0$ .

But $dV = -\vec{E} \cdot d\vec{r}$ . Setting this to zero means $\vec{E} \perp d\vec{r}$ — that is, the electric field is perpendicular to the surface at every point.

Mathematically: $\vec{E} = -\nabla V$ . The field points in the direction of steepest decrease of potential.

Equipotentials of common charge configurations

Point charge. $V = kq/r$ . Equipotentials are concentric spheres centred on the charge. Field lines are radial.

Uniform field (between parallel plates). $V$ varies linearly with distance from one plate. Equipotentials are parallel planes, evenly spaced (for a uniform field). Field lines are perpendicular to plates, parallel to each other.

Electric dipole. Equipotentials are complicated curves; the equatorial plane is an equipotential at $V = 0$ . Field lines emerge from $+q$ and end at $-q$ .

Conductor in electrostatic equilibrium. Entire conductor is one equipotential. The field inside is zero; outside, it’s perpendicular to the surface.

In one dimension: $E_x = -\dfrac{dV}{dx}$

In three dimensions: $\vec{E} = -\nabla V = -\left(\dfrac{\partial V}{\partial x}\hat{i} + \dfrac{\partial V}{\partial y}\hat{j} + \dfrac{\partial V}{\partial z}\hat{k}\right)$

Equipotential ⊥ field line, always.

Density of field lines ∝ field strength.

Equipotentials closely spaced ⇒ strong field there.

Properties of equipotential surfaces

They never intersect (else two different $V$ values at one point — contradiction).
They are perpendicular to field lines.
The work done to move a charge along an equipotential surface is zero.
Closer spacing of equipotentials means a stronger field there.
Inside a conductor at electrostatic equilibrium, the entire volume is at the same $V$ , so the conductor surface is an equipotential.

Solved Examples

Example 1 (CBSE) — Sketching equipotentials

Question. A positive point charge of $+5\ \mu$ C is at the origin. Sketch the equipotential surfaces at $V = 9{,}000$ V and $V = 4{,}500$ V.

Solution. $V = kq/r$ , so $r = kq/V = (9 \times 10^9 \times 5 \times 10^{-6})/V = 4.5 \times 10^4 / V$ .

For $V = 9000$ : $r = 5$ m. For $V = 4500$ : $r = 10$ m. Two concentric spheres of radii $5$ m and $10$ m. The outer one is at lower potential.

Example 2 (JEE Main) — Field from potential

Question. The potential in a region is $V(x, y) = 3x^2 + 2y$ (in volts, with $x, y$ in metres). Find the electric field at $(1, 2)$ .

Solution. $E_x = -\partial V/\partial x = -6x = -6$ V/m at $x = 1$ . $E_y = -\partial V/\partial y = -2$ V/m. So $\vec{E} = -6\hat{i} - 2\hat{j}$ V/m, magnitude $\sqrt{40} \approx 6.32$ V/m.

Example 3 (JEE Advanced) — Work along an equipotential

Question. A charge $+2\ \mu$ C is moved from point A to point B along an equipotential surface in an external field. The path is curved and $5$ m long. How much work is done?

Solution. Zero. By definition, $V$ is constant on an equipotential, so $\Delta V = 0$ , and $W = q\Delta V = 0$ . The path length is irrelevant.

Exam-Specific Tips

JEE Main weightage. Electrostatics contributes 4-5 marks per year. Equipotential surface questions appear as 1-mark conceptual MCQs (sketch / property based) and 4-mark numerical (find $\vec{E}$ from given $V(x,y,z)$ ).

NEET weightage. 1-2 questions in electrostatics. Equipotential is usually a single conceptual question.

CBSE. “Define equipotential surface; state two properties; draw equipotential surfaces for a uniform field” is a classic 3-mark question.

When you see “find the electric field at a point given $V(x,y,z)$ ,” instantly take partial derivatives. Don’t try to integrate or use Coulomb’s law. $\vec{E} = -\nabla V$ is faster.

Common Mistakes to Avoid

Drawing field lines parallel to equipotentials. They are always perpendicular. If your sketch shows them parallel, redraw immediately.
Treating zero potential as zero field. $V = 0$ at a point doesn’t mean $\vec{E} = 0$ there. The field depends on the gradient of $V$ , not its value. Example: midpoint between two equal and opposite charges has $V = 0$ but $\vec{E} \neq 0$ .
Equating “constant $V$ ” with “no force.” A test charge on an equipotential surface still feels a force (perpendicular to the surface). It just doesn’t gain or lose energy as it moves along the surface.
Thinking equipotentials must be planes. They’re planes only for a uniform field. For point charges, they’re spheres. For a dipole, they’re complex curves.
Confusing potential with potential energy. $V$ is potential (J/C). $U = qV$ is potential energy (J). Work done = $q\Delta V = \Delta U$ .

Practice Questions

Q1. Sketch equipotentials for two equal positive charges separated by distance $d$ .

Near each charge, equipotentials are nearly spherical. Far away, they look like equipotentials of a single $+2q$ charge. In between, they form a “figure-8” or peanut shape. The midpoint between the charges is a saddle point.

Q2. $V(x) = 5x^2 - 3x$ in volts, $x$ in metres. Find $E$ at $x = 2$ .

$E = -dV/dx = -(10x - 3) = -10(2) + 3 = -17$ V/m. So field is $17$ V/m in the $-x$ direction.

Q3. Show that work done to move a $1$ C charge from one face of a conductor to another (both inside) is zero.

A conductor in electrostatic equilibrium is one big equipotential. So $\Delta V = 0$ between any two points inside. Work $= q\Delta V = 0$ .

Q4. Why are equipotentials closer together in regions of strong field?

Because $E = -dV/dx$ . A large $E$ means $V$ changes rapidly with $x$ , so successive equipotentials (drawn at fixed $\Delta V$ ) are physically close to each other.

Q5. Two equipotential surfaces have $V_1 = 100$ V and $V_2 = 80$ V, separated by $0.05$ m of perpendicular distance. Estimate $\vec{E}$ .

$|E| \approx |\Delta V|/\Delta x = 20/0.05 = 400$ V/m, directed from the higher potential surface to the lower (i.e., from $V_1$ towards $V_2$ ).

Q6. True or false: equipotential surfaces of a dipole include the perpendicular bisector plane of the dipole axis.

True. On the perpendicular bisector, contributions from $+q$ and $-q$ are equal and opposite, so $V = 0$ everywhere on that plane.

Q7. What is the shape of equipotentials around an infinitely long uniformly charged wire?

Coaxial cylinders, with the wire as axis.

Q8. Why can equipotentials never intersect?

If they did, the potential at the intersection point would have two different values, which is impossible — $V$ is single-valued.

FAQs

Are equipotential surfaces real physical surfaces? No, they’re mathematical constructs — surfaces of constant potential. They help visualise the field structure.

Why is the surface of a conductor an equipotential? Because in electrostatic equilibrium, no current flows. If parts of the surface had different $V$ , charges would flow until they didn’t.

Can equipotentials be drawn for time-varying fields? With caution. For non-static cases, $V$ alone doesn’t determine $\vec{E}$ — you also need $-\partial \vec{A}/\partial t$ . JEE/NEET stick to electrostatics where this isn’t an issue.

Is the potential always zero at infinity? It’s a convention. We choose $V(\infty) = 0$ for finite charge distributions. For an infinite plane or wire, this convention fails and a different reference is chosen.

How do equipotentials look near a sharp edge of a conductor? They get crowded — meaning the field is very strong at sharp points. This is why lightning rods are pointed.

Can a moving charge cross an equipotential surface? Yes — it just gains or loses kinetic energy in proportion to the potential difference traversed.