Electrostatics problem types — Coulomb, Gauss, potential, capacitor classification

Question

A solid conducting sphere of radius $R$ carries a charge $Q$ . Find the electric field and potential at (a) a point outside the sphere ( $r > R$ ), (b) on the surface ( $r = R$ ), and (c) inside the sphere ( $r < R$ ).

(JEE Main / NEET — Electrostatics)

Electrostatics Problem Decision Tree

flowchart TD
    A["Electrostatics Problem"] --> B{What to find?}
    B -->|Force between charges| C["Coulomb's Law: F = kq1q2/r²"]
    B -->|Electric field| D{Symmetry present?}
    B -->|Potential| E["V = kQ/r or integrate E.dr"]
    B -->|Capacitance| F["C = Q/V or geometry formula"]
    D -->|Yes: spherical, cylindrical, planar| G["Gauss's Law: flux = Q_enc/epsilon_0"]
    D -->|No symmetry| H["Coulomb + Superposition"]
    G --> I["Choose Gaussian surface matching symmetry"]
    F --> F1["Parallel plate: C = epsilon_0 A/d"]
    F --> F2["Spherical: C = 4pi epsilon_0 ab/(b-a)"]

Solution — Step by Step

A conducting sphere has perfect spherical symmetry. We choose a spherical Gaussian surface concentric with the sphere.

Gauss’s law: $\oint \vec{E} \cdot d\vec{A} = \dfrac{Q_{\text{enc}}}{\varepsilon_0}$

By symmetry, $E$ is constant over the Gaussian surface and directed radially. So: $E \times 4\pi r^2 = Q_{\text{enc}}/\varepsilon_0$

The Gaussian surface encloses the entire charge $Q$:

E = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{kQ}{r^2}

This is identical to a point charge at the centre — the sphere behaves as if all charge is concentrated at its centre.

Potential: $V = \dfrac{kQ}{r}$

On the surface ( $r = R$ ):

E_{\text{surface}} = \frac{kQ}{R^2}, \quad V_{\text{surface}} = \frac{kQ}{R}

Inside ( $r < R$ ):

For a conductor, all charge resides on the surface. The Gaussian surface inside encloses zero charge.

\boxed{E_{\text{inside}} = 0, \quad V_{\text{inside}} = \frac{kQ}{R} \text{ (constant, same as surface)}}

The electric field is zero inside, but the potential is NOT zero — it equals the surface potential everywhere inside. Zero field means constant potential, not zero potential.

Why This Works

Gauss’s law relates the total electric flux through a closed surface to the enclosed charge. For symmetric charge distributions, we can pull $E$ out of the integral because it is constant over the Gaussian surface. This converts a vector calculus problem into simple algebra.

For conductors, free electrons redistribute until the internal field is zero — any non-zero field would move electrons, contradicting equilibrium. This self-shielding property makes conductors special.

Alternative Method — Using the Potential Approach

Instead of finding $E$ first, we can compute potential by integration:

V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vec{r}

For $r > R$ : $V = -\int_{\infty}^{r} \frac{kQ}{r^2}dr = \frac{kQ}{r}$

For $r < R$ : Since $E = 0$ inside, $V$ does not change from the surface value: $V = kQ/R$ .

JEE Advanced frequently combines electrostatics with other topics: a charged ball rolling (rotation + electrostatics), or a charged particle in a magnetic field (electrostatics + magnetism). Master each topic individually first, then practise combination problems from previous year papers.

Common Mistake

The most dangerous misconception: “Electric field is zero inside, so potential is also zero.” Wrong. Zero field means the potential is constant (no change), not zero. Inside a conductor, $V$ equals the surface potential. Confusing “constant” with “zero” leads to completely wrong energy calculations.