Question

A cube of side \( b \) has a charge \( q \) at seven of its vertices. The electric field due to this charge distribution at the centre of this cube will be:

(a) \( \frac{4 k \cdot q}{3 b^{2}} \) 

(b) \( k q / 2 b^{2} \) 

(c) \( 32 kq / b ^{2} \) 

(d) zero

Answer 1

4k.q/3b^2

the electric field due to other 6 vertices are cancel each other due to symmetry but the last electric field is :-

kq/(√3/2)b^2. (distance is the half of the diagonal =√3/2a)

Answer 2

The correct option is (a) \(\cfrac{4Kq}{3b^2}\)

Given side of cube is b then diagonal length of cube is r = \(\cfrac{\sqrt3b}2\)

the electric field AB will be effective and other will be cancle out

E = \(\cfrac{Kq}{r^2}\)

\(\because\) r = \(\cfrac{\sqrt3b}2\)

E = \(\cfrac{Kq}{(\cfrac{\sqrt3}2b)^2}\)

E = \(\cfrac{4Kq}{3b^2}\)