Question

A small sphere is charged uniformly and placed at point A(u, v) so that at point B(8, 7) electric field

strength is vector E = (54 i + 72j ) NC–1 and potential is + 900 volt. Calculate

(i) magnitude of charge,

(ii) co-ordinates of point A, and

(iii) if di-electric strength of air is 3 × 106 Vm–1, minimum possible radius of the sphere.

Answer 1

Since, potential due to sphere is positive, therefore, it is positively charged. Let magnitude of charge on sphere be q and let distance AB be equal to r.

Dividing equation (i) by (ii) r = 10 m

Substituting this value in equation (i),

q = 10–6 coulomb or 1µC

Since q is a positive charge, therefore

Since, minimum radius of sphere corresponds to electric field strength at surface of sphere to be equal to dielectric strength of air. Therefore radius R of sphere is given by,