Question

Surface density of charge on a sphere of radius 'R' in terms of electric intensity 'E' at a distance 'r' in free space is ($\epsilon_0$ = permittivity of free space)

• A. $\epsilon_0 E (\frac{R}{r})^2$
• B. $\frac{\epsilon_0 ER}{r^2}$
• C. $\epsilon_0 E (\frac{r}{R})^2$
• D. $\frac{\epsilon_0 Er}{R^2}$

Answer 1

Answer: (C)

$\epsilon_0 E \left(\frac{r}{R}\right)^2$

Answer 2

The surface charge density $\sigma$ is given by:

$$\sigma = \frac{Q}{4\pi R^2}$$

Using Gauss’s law:

$$E(4\pi r^2) = \frac{Q}{\epsilon_0}$$

Thus,

$$Q = 4\pi \epsilon_0 E r^2$$

Substituting $Q$ into the expression for $\sigma$:

$$\sigma = \frac{4\pi \epsilon_0 E r^2}{4\pi R^2} = \epsilon_0 E \left(\frac{r}{R}\right)^2$$