Question

how much work is required to move a charge of 4nC from a point 2m away to a point 0.5m away from a point charge of 60nC?what is the potential difference between these points?

Answer 1

The potential at point 2m away is-

$V_1=\dfrac{1}{4\pi \epsilon_o}\dfrac{q}{r}=9\times 10^9\times \dfrac{60\times 10^{-9}}{2}=270 {\text{volts}}$

The potential at point 0.5m away is-

$V_2=\dfrac{1}{4\pi \epsilon_o}\dfrac{q}{r}=9\times 10^9\times \dfrac{60\times 10^{-9}}{0.5}=1080 {\text{volts}}$

Then The potential difference between these points is-

$V=V_2-V_1=1080-270=810 \text{volts}$

Also, Work done required $=4\times 10^{-9}\times V=4\times 10^{-9}\times 810=3.24\times 10^{-6} J$