Q.A spherical conductor of radius $12\ \text{cm}$ has a charge of $1.6 \times 10^{-7}\ \text{C}$ distributed uniformly on its surface. What is the electric field

Concept: Electric Potential & Field of a Spherical Conductor

For a charged spherical conductor, the field inside is zero (electrostatic shielding). Outside, it behaves as if all charge is at the centre: $E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2}$.

Step 1 – Inside the sphere

For $r < R$, field is zero.

$\boxed{E_{\text{inside}} = 0}$.

Step 2 – Just outside the sphere

At $r = R = 0.12\ \text{m}$,

$E = \frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.12)^2}$

$= \frac{1440}{0.0144} = 1.0 \times 10^5\ \text{N/C}$.

Step 3 – At $18\ \text{cm}$ from centre …

The key to this problem is understanding electrostatic equilibrium in conductors. When a conductor is charged, the charges repel each other and move to the surface. Inside the conductor, the net electric field must be zero — otherwise, charges would keep moving. This is a fundamental property: the electric field inside a conductor in electrostatics is always zero.

For points outside the sphere, the charge on the spherical surface behaves exactly as if it were all concentrated at the centre. This is a consequence of Gauss's law and spherical symmetry — the field at a distance $r$ from the centre (for $r \ge R$) is the same as that of a point charge $Q$ at the centre.

Let's apply this step by step.

1. Inside the sphere ($r < R$)

The sphere is a conductor. In electrostatic equilibrium, the electric field inside the bulk of a conductor is zero. Since the charge resides only on the surface, any point inside (including the centre) experiences no net field.

So for part (a), the answer is simply $0$.

2. Just outside the sphere ($r = R$)

Here $r = 12\ \text{cm} = 0.12\ \text{m}$. For a point on the surface, we treat the sphere as a point charge at the centre. The electric field magnitude is:

We know $Q = 1.6 \times 10^{-7}\ \text{C}$ and $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \text{N m}^2/\text{C}^2$.

First compute the denominator: $(0.12)^2 = 0.0144$.

Numerator: $9 \times 10^9 \times 1.6 \times 10^{-7} = 14.4 \times 10^2 = 1440$.

So $E = \frac{1440}{0.0144} = 100000\ \text{N/C}$.

Thus $E = 1.0 \times 10^5\ \text{N/C}$, directed radially outward (since the charge is positive).

Method: Electric Field of a Uniformly Charged Conducting Sphere

This method finds the field at any point inside, on, or outside a charged spherical conductor by combining the conductor's electrostatic-equilibrium property with Gauss's law.

Steps

Step 1: Field inside a conductor is always zero

In electrostatic equilibrium, all free charge on a conductor resides on its outer surface, and the field inside the bulk material is zero — otherwise charge would still be migrating. This holds for every point with $r < R$, with no calculation needed.

Step 2: For any point at or outside the surface, replace the sphere with a point charge at its centre

By spherical symmetry and Gauss's law, a uniformly charged sphere produces exactly the same field outside itself as a point charge $Q$ placed at the centre: …