Source: PHY 212 Electromagnetism Practice Exam, University of Illinois at Urbana-Champaign
Difficulty: Intermediate | Prerequisites: Gauss's Law, Coulomb's Law, basic calculus (derivatives, line integrals)
Tags: electric field, electric potential, voltage, potential difference, equipotential lines, conductor, electrostatic equilibrium, grounding, shielding, Faraday cage, induced charge, gradient of potential, E from V
Electric potential gives you a scalar way to describe what the electric field is doing, which is often far easier to work with than the vector field itself. This set of notes ties together three ideas that exams love to combine: how the electric field behaves inside and on conductors at equilibrium, what electric potential means physically, and how E and V relate to each other mathematically. You should already be comfortable with Gauss's Law and with the idea that work and energy are scalars before diving in. If you skipped the Gauss's Law notes, go back, because the conductor arguments here lean on it.
The electric field inside a conductor in electrostatic equilibrium is always zero, charges sit on the surface, and the whole conductor is at a single potential. Electric potential is the work per unit charge done by an external agent to move a test charge from a reference point, and the field is recovered from the potential by taking the negative gradient. Equipotential surfaces are always perpendicular to field lines.
Electric potential (V)
The electric potential at a point is the work done per unit positive test charge by an external agent in bringing the charge from a chosen reference point to that point, without acceleration. In simple terms, it is the "electrical height" at a location: charges roll downhill from high V to low V, just as balls roll downhill under gravity.
Potential difference (Delta V, or V_a - V_b)
The difference in electric potential between two points. It equals the negative of the line integral of E dot dl from a to b. Think of it as the voltage your voltmeter reads between two probes.
Equipotential line (or surface)
A curve (or surface in 3D) along which the electric potential is constant. No work is done moving a charge along an equipotential. Equipotential surfaces are always perpendicular to electric field lines.
Electrostatic equilibrium
The state in which all charges in a conductor have stopped moving. In this state, the electric field inside the conductor is zero, any excess charge resides entirely on the outer surface, and the entire conductor is at one potential.
Grounding
Connecting a conductor to the Earth (or to a large reservoir of charge at V = 0). Grounding allows charge to flow on or off the conductor until its potential equals zero. In simple terms, the Earth acts as an infinite source or sink of charge.
Electrostatic shielding (Faraday cage effect)
The phenomenon whereby the electric field inside a hollow conductor is zero regardless of external fields. An external charge induces surface charges on the conductor that exactly cancel the external field inside.
Electrostatic induction
The redistribution of charge on a conductor caused by a nearby external charge. The side of the conductor nearest the external charge acquires an opposite sign, and the far side acquires the same sign. The conductor remains electrically neutral overall (unless grounded).
At electrostatic equilibrium, the electric field everywhere inside the conducting material is exactly zero: E = 0.
If E were not zero, free charges would move, contradicting the assumption of equilibrium.
Any excess charge resides entirely on the outer surface of the conductor.
The electric field just outside the surface is perpendicular to it, with magnitude sigma / epsilon_0, where sigma is the local surface charge density.
Consider a solid inner cylinder (radius a, linear charge density lambda_inner) surrounded by a cylindrical shell (inner radius b, outer radius c, total charge density lambda_shell).
Without grounding, the inner surface of the shell carries -lambda_inner (induced by the inner cylinder), and the outer surface carries lambda_shell + lambda_inner.
When the outer shell is grounded, its potential is forced to zero. Charge flows to or from the Earth to achieve this. The induced charge on the inner surface of the shell (at radius b) remains unchanged at -lambda_inner, because it is determined entirely by the charge on the inner cylinder via Gauss's Law. What changes is the charge on the outer surface.
For two coaxial cylinders with lambda_inner = lambda_shell = 0, there is no charge anywhere. The electric field between them is zero, and therefore the potential difference V_a - V_c = 0.
More generally, when lambda_inner is nonzero, V_a - V_c is found by integrating E dot dr from a to c using the cylindrical field E = lambda_inner / (2 pi epsilon_0 r).
Usually V is defined as zero at infinity. But exams sometimes set V = 0 at a different location.
Example: a uniformly charged insulating sphere of radius R with total positive charge Q. If V = 0 at the surface (r = R), then at point P (r = 2R, outside the sphere), the potential is found by integrating E from R to 2R.
Outside the sphere, E = kQ/r^2 pointing outward. Integrating from R to 2R: V(2R) - V(R) = -integral from R to 2R of (kQ/r^2) dr = -kQ(-1/r) from R to 2R = -kQ(1/2R - 1/R) = -kQ(-1/2R) = kQ/(2R).
Wait: V(R) = 0 by definition, so V(2R) = V(2R) - 0 = -kQ/(2R). The potential at P is negative, because you are moving in the direction the field pushes a positive charge (downhill in potential), so V decreases.
The key insight: for a positive charge, V decreases as you move outward from the surface, so defining V = 0 at the surface makes V negative at points farther away.
Inside a uniformly charged insulating sphere (r < R), apply Gauss's Law with a spherical Gaussian surface of radius r. The enclosed charge is Q(r/R)^3.
This gives E = kQr / R^3 inside the sphere. The field increases linearly from zero at the centre to kQ/R^2 at the surface.
At the centre (r = 0), the field is zero.
Equipotential lines are always perpendicular to electric field lines. This is because the field does no work on a charge moving along an equipotential (dV = 0 along that path, so E dot dl = 0, meaning E is perpendicular to dl).
If E = 0 in a region, then V is constant throughout that region (since dV = -E dot dl = 0 for any path). The converse also holds: constant V implies E = 0 in that region.
Critically, E = 0 does not mean V = 0. It means V is constant, but that constant could be any value.
The electric field is the negative gradient of the potential: E = -grad V = -(dV/dx, dV/dy, dV/dz).
Each component of E is the negative partial derivative of V with respect to that coordinate.
This is the standard method when V is given as a function and E is needed.
A hollow uncharged conducting shell shields its interior from external electric fields. If a charge is placed outside the shell, the field inside the hollow region is zero.
This works because the external charge induces surface charges on the conductor that rearrange to cancel the field inside. By Gauss's Law, a Gaussian surface inside the conductor encloses zero charge and E = 0 in the conductor, and by the uniqueness theorem, the field in the interior cavity is also zero.
An uncharged metal sphere brought close to a positively charged metal surface will be attracted to it.
The positive surface induces negative charge on the near side of the sphere and positive charge on the far side. Because the negative charges are closer, the attractive force is stronger than the repulsive force, and the net force is attractive.
The potential energy of two point charges q1 and q2 separated by distance r is U = kq1q2 / r.
For two charges of the same sign (both positive or both negative), U is positive. Doubling the separation to 2d halves the denominator, so U decreases by a factor of 2.
E inside a conductor at equilibrium: E = 0
E just outside a conductor: E = sigma / epsilon_0 (perpendicular to surface)
E from potential: E = -grad V = -(dV/dx, dV/dy, dV/dz)
Potential difference: V_a - V_b = -integral from a to b of E dot dl
E inside a uniformly charged sphere (r < R): E = kQr / R^3
Potential energy of two point charges: U = kq1q2 / r
Point charge potential: V = kQ / r (with V = 0 at infinity)
Electrostatic shielding is why sensitive electronics are housed in metal enclosures, and why your car (a metal shell) is a relatively safe place during a lightning storm. Grounding is a core safety principle in electrical engineering: connecting equipment casings to earth ensures that a fault current flows safely to ground rather than through a person.
Students often think that E = 0 implies V = 0. It does not. Zero field means constant potential, but that constant can be any value.
Students often think that grounding a conductor eliminates the induced charge on its inner surface. It does not. The inner surface charge is fixed by the enclosed charge via Gauss's Law. Grounding changes only the outer surface charge.
Students often think an uncharged conductor near a charged object feels no force. It does, because of electrostatic induction: the induced charges are at different distances from the external charge, producing a net attractive force.
Students often think the potential at a point must be positive if the source charge is positive. This depends entirely on the reference point. With a non-standard reference (e.g. V = 0 at the surface of a sphere), V can be negative at points farther away from a positive charge.
⚠️ The relationship E = 0 implies V = constant (not V = 0) is tested very frequently. Be precise.
⚠️ Expect a question with a non-standard reference point for V. Work through the integral carefully and mind the sign.
⚠️ Know that equipotential lines are perpendicular to field lines, not parallel or at 45 degrees.
⚠️ Grounding questions almost always hinge on distinguishing which surface charge changes and which does not.
⚠️ The formula E = -dV/dx (or the full gradient) is a standard short-answer target.
True or false: The electric field inside a solid conductor at electrostatic equilibrium is always zero. (True.)
Fill in the blank: Equipotential surfaces are always __________ to electric field lines. (perpendicular)
True or false: If E = 0 in a region, the potential in that region must be zero. (False. It must be constant, but not necessarily zero.)
Fill in the blank: The electric field is the negative __________ of the potential. (gradient)
True or false: Two same-sign point charges that are moved farther apart have higher potential energy. (False. U = kq1q2/r decreases as r increases for same-sign charges.)
Q: If we define V = 0 at the surface of a uniformly charged positive insulating sphere (r = R), what is the sign of V at point P (r = 2R)?
A: Negative. The field points outward, so moving from R to 2R is moving in the direction of E. Potential decreases in the direction of E, so V(2R) < V(R) = 0.
Q: What is the electric field magnitude inside a solid conductor at electrostatic equilibrium?
A: E = 0. Free charges redistribute until the internal field vanishes.
Q: If the distance between two equal same-sign point charges is doubled from d to 2d, what happens to the electric potential energy?
A: It decreases by a factor of 2. U = kq^2/r, so doubling r halves U.
Q: If the electric field is zero in a region, what can be concluded about the potential?
A: The potential must be constant throughout that region. It need not be zero.
Q: Briefly explain why the electric field inside a hollow uncharged conducting shell is zero when a charge is placed outside.
A: The external charge induces surface charges on the conductor. These rearrange so that, by Gauss's Law, the net field inside the conducting material is zero. By the uniqueness theorem, the field in the interior cavity is also zero. The conductor shields its interior.
Q: How do you calculate the electric field if V(x, y, z) is known?
A: Take the negative gradient: E = -(dV/dx, dV/dy, dV/dz). Each component of E is the negative partial derivative of V with respect to that spatial coordinate.
Electric potential connects directly to capacitance (see the Capacitors notes): capacitance is defined as Q/V, and the energy stored in a capacitor is expressed in terms of V. The gradient relationship E = -grad V also appears in the study of Laplace's equation and boundary-value problems later in the course. The conductor-at-equilibrium rules reappear whenever you analyse circuits, because wires are conductors and the potential is the same everywhere along an ideal wire.
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