Question

Two charged conducting spheres of radii 5 cm and 10 cm have equal surface charge densities. If the electric field on the surface of the smaller sphere is E, then the electric field on the surface of the larger sphere is

• A. 2E
• B. 4E
• C. 0.5E
• D. E

Hint:

- Electric field on the surface of a charged conducting sphere of radius R and charge Q: \( E = \frac{kQ}{R^2} = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2} \). - Surface charge density: \( \sigma = \frac{Q}{\text{Surface Area}} = \frac{Q}{4\pi R^2} \). - From these, \( E = \frac{\sigma (4\pi R^2)}{4\pi\epsilon_0 R^2} = \frac{\sigma}{\epsilon_0} \). - If \( \sigma \) is the same for two spheres, then \(E\) on their surfaces is the same, regardless of their radii.

Answer 1

The Correct Option is D

Let \( \sigma \) be the surface charge density.
Given it's equal for both spheres.
For a conducting sphere of radius R and charge Q, the surface charge density is \( \sigma = \frac{Q}{4\pi R^2} \).
The electric field E on the surface of a charged conducting sphere is \( E = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2} \).
Substitute \( Q = \sigma (4\pi R^2) \) into the electric field formula: \[ E = \frac{1}{4\pi\epsilon_0} \frac{\sigma (4\pi R^2)}{R^2} = \frac{\sigma}{\epsilon_0} \] This shows that the electric field on the surface of a conducting sphere depends only on its surface charge density \( \sigma \) and the permittivity of free space \( \epsilon_0 \).
Since both spheres have equal surface charge densities \( \sigma \), the electric field on their surfaces will be the same.
Given that the electric field on the surface of the smaller sphere is E.
Therefore, the electric field on the surface of the larger sphere is also E.
This matches option (4).
The radii \(R_1 = 5\) cm and \(R_2 = 10\) cm are not needed if surface charge densities are equal.
If charges were equal, then \( E \propto 1/R^2 \).
If potentials were equal, then \( V = \frac{Q}{4\pi\epsilon_0 R} \).
If \(V\) is same, \(Q \propto R\), then \( \sigma = Q/(4\pi R^2) \propto R/R^2 = 1/R \).
Then \( E = \sigma/\epsilon_0 \propto 1/R \).