Question

1 mC charge is placed on each vertex of a regular hexagon. Side of hexagon is 1 m, then electric field at

its centre is ...... .
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Answer 1

Given:

1 mC charge is placed on each vertex of a regular hexagon of side of hexagon is 1 m.

To find:

Net electrostatic field intensity at the centre of the hexagon.

Calculation:

The electrostatic field intensity contributed by individual charges at each vertex of the hexagon will be equal to :

$\therefore \: E = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{ \dfrac{1 \times {10}^{ - 3} }{1} \bigg \}$

$= > \: E = \dfrac{ {10}^{ - 3} }{4\pi\epsilon_{0}}$

$= > \: E = {10}^{9} \times {10}^{ - 3}$

$= > \: E = {10}^{6} \: N{C}^{-1}$

Now , the net field intensity at the centre of the hexagon will be zero because the field intensity of one charge will be cancelled by the diametrically opposite charge.

$\therefore \: E_{net}= 0\: N{C}^{-1}$

So, final answer is:

$\boxed{ \bf{ \: E_{net}= 0\: N{C}^{-1}}}$

Answer 2

Answer: Zero

Explanation:

As it is a regular hexagon and all the vertices have equal charge the electric field due to one charge will be cancelled out by the charge opposite to it, the net electric field at the centre will be zero.