Question

eight identical point charges of q coulomb each are placed at corners of a cube of each side 0.1m. Calculate electric field at the centre of the cube. Calculate the field at centre when one of the corner charges is removed

Answer 1

as the 8 charges are symmetrically placed, the electric field due to each of the four charges will be cancelled due to the field from the four charges located in the diagonally opposite corner...

if one charge is removed, then the resultant is equal in magnitude and opposite in sign to the field contributed by that charge (removed one)...

so   1/(4πε)  *  q/(√3 a)²     
        a = 0.1 m
   E = 3 * 10¹¹ * q    N/C

Answer 2

Answer:

ELECTRIC FIELD AT THE CENTRE WILL BE ZERO this is because diagonally opposite charges produces equal and opposite field which cancel out each other.

Explanation:

now the field at centre when one of the corner charges is removed can be given by

E=kq/r^2

the distance between centre and the vertices of the cube is a√3/2 where a= 0.1m

therefore

E=9×10^9q/(0.1√3)^2= 1.2 ×10^11 N/C