Question

A.) 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?

B.) What is the potential difference between these points?

Answer 1

The potential of 60nC-charge at a distance 2 and 0.m away correspondingly:

$$V_2=\frac{kq}{r_2}=270\text{ V},\\\space\\ V_{0.5}=\frac{kq}{r_{0.5}}=1080\text{ V}.$$

A) The work:

$$W=Q\Delta V=(4·10^{-9})(1080-270)=\\=3.24·10^{-6}\text{ J}.$$

B) The potential difference:

$$\Delta V=V_{0.5}-V_2=1080-270=810\text{ V}.$$