Question

six point charges are placed at the vertices of a hexagon of side 1m . Net electric field at the centre of the hexagon is


q______q

/

q-______/q

-q -q

Answer 1

Answer:

The electric field at the centre will be 0.

Explanation:

We consider hexagon with charge ‘’q’’ at each corner and we find the force due to these charges at center O.

Magnitude of electric field due to each charge at centre of hexagon will be equal to the sum of charges due to all segments.

For segment BE

$F_{B E}=\frac{k(q)(-q)}{1^{2}}=\frac{-k q^{2}}{1}$

For segment AD

$F_{A D}=\frac{k(-q)(q)}{1^{2}}=-k q^{2}$

For segment FC

$F_{F C}=\frac{k(-q)(+q)}{1^{2}}=-k q^{2}$

Total charge at center O is $F_{B E}+F_{A D}+F_{F C}=-k q^{2}-k q^{2}-k q^{2}=-3 k q^{2}$

The negative sign of the total force at the center denotes the property of the charge that will be acquired with respect to other charges.