Question

A charge Q is given to a neutral conducting plate of area A as shown below. The net electric field at point P which is located very near to the plate as shown in the figure is:

• A. $\frac{Q}{A\epsilon_0}$
• B. $\frac{Q}{2A\epsilon_0}$
• C. $\frac{Q}{4A\epsilon_0}$
• D. $\frac{2Q}{A\epsilon_0}$

Answer 1

Answer: (B)

$\frac{Q}{2A\epsilon_0}$

Answer 2

When a charge $Q$ is given to an isolated, thin conducting plate, the charges distribute equally on its two faces (assuming symmetry), so each face gets a charge of $Q/2$. The electric field just outside a conductor is given by

$$E = \frac{\sigma}{\epsilon_0},$$

where the surface charge density is

$$\sigma = \frac{Q/2}{A} = \frac{Q}{2A}.$$

Thus, the field at point $P$ (which is near the plate on one side) is

$$E = \frac{Q}{2A\epsilon_0}.$$