Question

a conducting surface of 5cm is chargee to 15microcm​

Answer 1

Let common potential = V

q1 = 4πεor1 V , q2 = 4πεor2 V

σ1 = q1/4πr12 ; σ2 = q2/4πr22

σ1/σ2 = (q1/q2) (r22/r12) = r2/r1 = 10/5 = 2 : 1