Electric potential (V) due to a point charge (q) at a distance (r) is given by:
$V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$
Given: $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{N m}^2 \, \text{C}^{-2}$ q = 4 × 10-7 C r = 9 cm = 0.09 m
$V = (9 \times 10^9 \, \text{N m}^2 \, \text{C}^{-2}) \frac{(4 \times 10^{-7} \, \text{C})}{0.09 \, \text{m}} = 4 \times 10^4 \, \text{V}$
AB is a part of an electrical circuit (see figure). The potential difference \(V_A - V_B\), at the instant when current \(i = 2\) A and is increasing at a rate of 1 amp/second is: