Question

5 point charges each of charge +q are placed on five vertices of a regular hexagon of side l .Find the magnitude of resultant force on charge -q placed at centre of hexagon

Answer 1

Answer:

Force on the charge placed at the center of the hexagon is given as

$F = \frac{kq^2}{l^2}$

Explanation:

As we know that 5 point charges are placed at five vertices of regular hexagon

So here net force on the charge placed at the center of the hexagon is to be find out

As we know that if all charges are identical and placed at all vertices of the hexagon then net electric field at the center must be ZERO

so here we can say

$E_1 + E_2 + E_3 + E_4 + E_5 + E_6 = 0$

now here we need to find the electric field due to 5 point charges so by super position law we can say

$E_1 + E_2 + E_3 + E_4 + E_5 = - E_6$

so we have electric field at the center is equivalent to the field of one charge only

$E = \frac{kq}{l^2}$

now force on the charge placed at the center is given as

$F = \frac{kq^2}{l^2}$

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Topic : Electrostatic Force

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