Question

A neutral conducting solid sphere of radius 2 m is placed such that its centre is 4 m from a point charge q=2mC. The electric field intensity just outside a point A on the sphere is found to be 2 N/C. If the charges per unit area at A is xε 0 ​ , find the value of x.

Answer 1

x = 2

Answer 2

1. For a conductor in electrostatic equilibrium, the electric field just outside (normal to the surface) is related to the surface charge density by

   $$E = \frac{\sigma}{\varepsilon_0}$$

   where $\sigma$ is the surface charge density.

2. The problem states that at point A the electric field is $2\,\text{N/C}$. Thus,

   $$\sigma = \varepsilon_0 \times E = \varepsilon_0 \times 2 = 2\varepsilon_0.$$

3. Since the charge per unit area at A is given as $x\,\varepsilon_0$, we immediately have

   $$x = 2.$$

Minimal Explanation:

Using the conductor boundary condition,

$$\sigma = \varepsilon_0 E$$

and given $E = 2\,\text{N/C}$, we obtain $\sigma = 2\varepsilon_0$. Hence, $x = 2$.