Question

What is the potential difference between a point 0.48 mm from a charge of 2.9 nc and a point at infinity?

Answer 1

The potential at a distance r from a charge Q is given by
$V(r) = k_e \frac{Q}{r}$
where ke is the Coulomb's constant.

The charge in our problem is $Q=2.9 nC=2.9 \cdot 10^{-9} C$; for the point at $r=0.48 mm=0.48 \cdot 10^{-3} m$, the potential is
$V_1 = k_e \frac{Q}{r}= (8.99 \cdot 10^9 Nm^2 C^{-2}) \frac{2.9 \cdot 10^{-9} C}{0.48 \cdot 10^{-3} m}= 5.43 \cdot 10^4 V$

For the point at infinity, we immediately see that the potential is zero, because $r= \infty$ and so $V_2 = 0$.

Therefore, the potential difference between the two points is
$\Delta V = V_1 - V_2 = V_1 = 5.43 \cdot 10^4 V$