Difficulty: Intermediate | Prerequisites: Basic calculus (integrals, derivatives), vector notation, Coulomb's Law fundamentals
Electrostatics is the foundation of the entire Electricity & Magnetism course. Everything here, from how charges distribute on conductors to how Gauss's Law simplifies field calculations, feeds directly into capacitance, circuits, and electromagnetic theory later on. If you are coming in cold, make sure you are comfortable with superposition of forces, the idea of a field, and basic integration before diving in. This material accounts for a significant fraction of a typical E&M final.
Electric charges create electric fields, and those fields exert forces on other charges. Gauss's Law lets you calculate the electric field for symmetric charge distributions by relating the total flux through a closed surface to the enclosed charge. The electric potential is the scalar counterpart of the field: where potential is constant, the field is zero, and no work is done moving charges along equipotential surfaces.
Electrostatic induction
The redistribution of charge on a neutral conductor caused by a nearby charged object, without any physical contact. Think of it as: the nearby charge "pushes" same-sign charges to the far side and "pulls" opposite-sign charges to the near side.
Gaussian surface
An imaginary closed surface used in Gauss's Law to calculate electric flux. In simple terms, this means you pick a shape (sphere, cylinder) that matches the symmetry of the charge distribution so the maths simplifies.
Electric flux
The total "flow" of electric field through a surface, calculated as the surface integral of E dot dA. Think of it as: how many field lines pass through a given area.
Electrostatic equilibrium
The state of a conductor in which all charges have stopped moving. In simple terms, this means the electric field inside the conductor is zero and all excess charge sits on the surface.
Equipotential surface
A surface on which the electric potential has the same value everywhere. Think of it as: a contour line on a topographic map, but for voltage instead of altitude. No work is done moving a charge along one.
Electric potential (V)
The electric potential energy per unit charge at a point in space, measured in volts. In simple terms, this means it tells you how much energy a positive test charge would have at that location, per coulomb of charge.
Volume charge density (ρ)
The amount of charge per unit volume, measured in C/m³. Think of it as: how densely charge is packed into a region of space.
Conducting shell
A hollow conductor, often spherical. Charge placed inside it induces equal and opposite charge on the inner surface, with the remainder appearing on the outer surface.
When an uncharged metal object is brought near a charged object, charges in the metal redistribute.
The side nearest the charged object acquires the opposite sign of charge (attracted).
The far side acquires the same sign (repelled).
The net charge on the metal remains zero, but because the attracted charges are closer, the net force is attractive.
This is why an uncharged metal sphere brought near a negatively charged surface is attracted to it, even though the sphere has no net charge.
Gauss's Law states that the total electric flux through any closed surface equals the enclosed charge divided by ε₀.
The law is always true, but it is only useful for direct calculation when the charge distribution has enough symmetry (spherical, cylindrical, or planar) that E can be pulled out of the integral.
For a charged solid cube, there is no Gaussian surface that exploits symmetry. Gauss's Law cannot be used to easily calculate the field. You would need direct integration instead.
For spherical charge distributions (shells, solid spheres), use a spherical Gaussian surface of radius r:
If r is inside a uniformly charged shell (inner radius a, outer radius b), the enclosed charge is only the charge within radius r.
The enclosed charge for a < r < b is ρ times the volume of the shell from a to r: Q_enc = ρ · (4/3)π(r³ – a³).
Apply Gauss's Law: E(4πr²) = Q_enc / ε₀, then solve for E.
For r < a (the hollow interior of an insulating shell with no charge inside), E = 0 by Gauss's Law.
Inside a conductor at electrostatic equilibrium, the electric field is zero. If it were not, free charges would move until it became zero.
All excess charge resides on the surface.
The electric field just outside the surface is perpendicular to the surface.
If a charge Q is placed at the centre of a neutral conducting shell, the inner surface acquires charge –Q (to make the field zero inside the conductor), and the outer surface acquires +Q (to keep the shell's net charge zero).
Inner shell: radius a, charge +Q. Outer shell: radius b, charge –Q.
Region r < a (inside the inner shell): E = 0, because the shell is a conductor.
Region a < r < b (between the shells): Apply Gauss's Law with a spherical surface of radius r. Enclosed charge = +Q. So E = Q / (4πε₀r²), directed radially outward.
Region r > b (outside both shells): Enclosed charge = +Q + (–Q) = 0. So E = 0.
The potential in the region a < r < b is found by integrating E inward from infinity. Since E = 0 for r > b, V(b) = 0 (because V = 0 at infinity and no field to integrate through). Then for a < r < b, V(r) = Q/(4πε₀) · (1/r – 1/b).
The electric field is the negative gradient of the potential: E = –dV/dr (in one dimension) or E = –∇V.
If V is constant throughout a region, then E = 0 in that region. A uniform potential means no field.
This is one of the most commonly tested relationships on an E&M exam.
The work done by the electric field in moving a charge along an equipotential surface is zero.
This follows directly from W = qΔV, and ΔV = 0 along an equipotential.
The field is always perpendicular to equipotential surfaces, so no component of force acts along the surface.
\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}Gauss's Law: the net electric flux through any closed surface equals the enclosed charge divided by the permittivity of free space.
\vec{E} = -\nabla VThe electric field is the negative gradient of the electric potential. In one dimension, E = -dV/dx.
W = q \Delta VWork done by the electric field on a charge q equals the charge times the change in potential. On an equipotential surface, ΔV = 0, so W = 0.
E = \frac{Q}{4\pi \varepsilon_0 r^2}Electric field at distance r from a point charge or outside a spherically symmetric distribution (r greater than the outer radius).
V(r) = \frac{Q}{4\pi \varepsilon_0} \left( \frac{1}{r} - \frac{1}{b} \right)Potential between concentric shells (a < r < b) with inner charge +Q and outer charge -Q, with V = 0 at infinity.
Students often think Gauss's Law only applies to symmetric charge distributions. It applies to all distributions; symmetry just makes the integral solvable by hand.
Students often assume that if the electric field is zero inside a conductor, there is no charge on the conductor. There can be plenty of charge, all on the surface.
Students confuse "constant potential" with "non-zero field." If V is the same everywhere in a region, E is zero there. A constant, non-zero potential does not produce a field.
Students sometimes believe that bringing an uncharged conductor near a charged object will repel it. The induced charge redistribution always produces a net attractive force.
Electrostatic induction is the principle behind how lightning rods work: the rod's sharp tip concentrates induced charge, creating a path for lightning to follow safely to ground. Gauss's Law is essential in designing the shielding of sensitive electronics (Faraday cages), where the zero internal field of a conductor protects the contents from external electric fields.
⚠️ The relationship E = –∇V (or E = 0 when V is constant) is tested in nearly every E&M exam. Know it cold.
⚠️ Gauss's Law questions often test whether you recognise when the law cannot be used to easily calculate a field (lack of symmetry, e.g. a cube).
⚠️ Conductors at equilibrium: expect a question on zero internal field and on how charge distributes when a charge is placed inside a conducting shell.
⚠️ Work along an equipotential is always zero. This is a quick-mark question that students lose points on by overthinking.
True or False: The electric field inside a conductor at electrostatic equilibrium is always zero. (True)
Fill in the blank: If the electric potential is constant throughout a region, the electric field in that region is __________. (Zero)
True or False: Gauss's Law can be used to easily calculate the electric field of a charged cube. (False, the cube lacks the symmetry needed.)
Fill in the blank: When a charge +Q is placed at the centre of a neutral conducting shell, the inner surface of the shell acquires a charge of __________. (–Q)
True or False: An uncharged metal sphere brought near a negatively charged surface will be repelled. (False, it will be attracted due to induction.)
Q: An uncharged metal sphere is brought near a negatively charged metal surface. What happens?
A: The sphere is attracted to the surface. Negative charges on the surface repel electrons in the sphere to the far side, leaving the near side positively charged. The attraction between the near positive charges and the surface's negative charges is stronger (shorter distance), so the net force is attractive.
Q: You need to find the electric field at a distance R from a charged solid cube of side a. Can you use Gauss's Law to calculate it directly?
A: No. A cube lacks the spherical, cylindrical, or planar symmetry needed for Gauss's Law to yield a simple, solvable integral. You would need to integrate Coulomb's Law contributions from each volume element of the cube.
Q: A charge +Q is placed at the centre of a neutral conducting spherical shell. What is the induced charge on the inner surface?
A: The induced charge on the inner surface is –Q. A Gaussian surface drawn within the conductor (where E = 0) must enclose zero net charge, so the inner surface must carry –Q to cancel the +Q at the centre.
Q: If the electric potential V is 50 V everywhere in a region, what is the electric field there?
A: The electric field is zero. E = –∇V, and if V is constant (no spatial variation), all partial derivatives are zero.
Q: A charged spherical insulating shell has inner radius a and outer radius b, with uniform volume charge density ρ. Derive E for r < a.
A: For r < a, draw a spherical Gaussian surface of radius r. The shell has no charge in the region r < a (all charge is between a and b), so Q_enc = 0 and E = 0.
Q: For the same shell, derive E in the region a < r < b.
A: The enclosed charge is the charge in the shell from radius a to r: Q_enc = ρ · (4/3)π(r³ – a³). By Gauss's Law, E · 4πr² = ρ(4/3)π(r³ – a³)/ε₀. So E = ρ(r³ – a³) / (3ε₀r²).
Q: Two concentric conducting shells have charges +Q (inner, radius a) and –Q (outer, radius b). What is E for r > b?
A: E = 0. The total enclosed charge is +Q + (–Q) = 0.
The electric potential concepts here lead directly into capacitance: a capacitor stores energy by maintaining a potential difference between two conductors. Gauss's Law for electric fields has a direct magnetic counterpart (Gauss's Law for magnetism), where the flux of B through any closed surface is always zero (no magnetic monopoles). The relationship E = –∇V is the electrostatic version of Faraday's Law, which generalises to time-varying fields and electromagnetic induction.
Coulomb's Law, electric field, electric flux, Gauss's Law, Gaussian surface, electrostatic induction, conductor, electrostatic equilibrium, electric potential, voltage, equipotential surface, potential difference, charge density, volume charge density, surface charge, conducting shell, concentric shells, Faraday cage, shielding, point charge field, superposition, permittivity of free space, ε₀, inverse square law, spherical symmetry, PHYS 212, University Physics E&M