Question

how to find potential in such cases?​

Answer 1

Answer:

9

Explanation:

Given coulomb constant (k) = 9 ×10^9

Charge (q) = 1 × 10^-9

Distance of separation (r) = 1m

Electric potential = V = k×q/r

V =9×10^9×1×10^-9/1

V = 9

The electric potential at centre of cavity is 9 volts