Question

Six point charges are placed at the vertices of a hexagon of side $1m$ as shown in figure. Net electric field at the centre of the hexagon is

• A. $Zero$
• B. $\frac{6q}{4\pi {ϵ}_0}$
• C. $\frac{q}{\pi {ϵ}_0}$
• D. $\frac{q}{4\pi {ϵ}_0}$

Answer 1

The correct option is C $\frac{q}{\pi {ϵ}_0}$

Each $△$ are equilateral triangle in symmetrical hexagon.

$E.F$ at each diagonal

$E=\frac{kq}{r^2}+\frac{kq}{r^2}=\frac{2kq}{1}=2kq$

Resultant $EF$ due to two diagonal at an angle ${60}^{\circ }$

$E.F_{net}=\sqrt{E^2+E^2+2E^2\cos {60}^{\circ }}$

$=\sqrt{{\left(2kq\right)}^2+{\left(2kq\right)}^2+{\left(2kq\right)}^2}$

$=2kq\sqrt{3}$

Now,

$E_{net}=\sqrt{{\left(2kq\sqrt{3}\right)}^2+{\left(2kq\right)}^2}$

$=2kq\sqrt{4}$

$=4kq$

$=\frac{4\times q}{4\pi {\epsilon }_0}$

$E_{net}=\frac{q}{\pi {\epsilon }_0}$

Hence option (C) is correct