Question

Five point charges $(+q$ each$)$ are placed at the five vertices of a regular hexagon of side $2a$. What is the magnitude of the net electric field at the centre of the hexagon?

• A. $\frac{1}{4\pi {ϵ}_0}\frac{q}{a^2}$
• B. $\frac{q}{16\pi {ϵ}_0{a}^2}$
• C. $\frac{\sqrt{2}q}{4\pi {ϵ}_0{a}^2}$
• D. $\frac{5q}{16\pi {ϵ}_0{q}^2}$

Answer 1

The correct option is B $\frac{q}{16\pi {ϵ}_0{a}^2}$

Five point charges ($+q$ each) are placed at the five vertices of a regular hexagon of side $2a$.

Let, the electric fields be as shown in the figure :


$E_1$ and $E_4$ are cancelled (equal in magnitude and opposite in direction)

Similarly, $E_2$ and $E_5$ are cancelled.

The net electric field will be due to the charge at $3$ only,

$\Rightarrow E_{\text{net}}=E_3$

$\Rightarrow E_{net}=\frac{1}{4\pi {ϵ}_0}\frac{q}{(2a{)}^2}$

$\therefore E_{net}=\frac{q}{16\pi {ϵ}_0{a}^2}$

Hence, option (b) is correct answer.