Question

A positive charge $q$ is placed in front of a conducting solid cube of side $a$ at distance $d$ from its center. Find the electric field at the center of the cube due to the charges appearing on its surface.

• A. Zero
• B. $\frac{2Kq}{d^2}$
• C. $\frac{Kq}{4d^2}$
• D. $\frac{Kq}{d^2}$

Answer 1

The correct option is D $\frac{Kq}{d^2}$

Since field inside a conductor is always zero.i.e., $E_{\text{conductor}}=0$

Hence, field due to charge $q$ and those appearing at surface should cancel each other.

So, we have
$E_{\text{conductor}}=E_{\text{surface charge}}+E_q=0.....\left(1\right)$

Where,
$E_{\text{surface charge}}=$Electric field at the centre of conductore due the charges appearing on the surface.

$E_q=\frac{Kq}{d^2}=$Electric field due the point charge at the centre of the conductor.

From equation $\left(1\right)$, we get
$E_{\text{surface charge}}=-E_q=-\frac{Kq}{d^2}$

Hence, the magnitude of field at center due to charges appearing at surfaces is $\frac{Kq}{d^2}$, directed towards the charge $q$.

Thus, option (b) is the correct answer.