Question

A point charge 'q' produce an absolute potential of 5 V at a distance of 1 meter. What will be the Absolute Potential due to 'q' at a distance of 3 meter ?

Answer 1

By definition, the absolute potential of point charge at the distance $r$ is:

$$\varphi = k\dfrac{q}{r}$$

where $k$ is a constant.

A point charge 'q' produce an absolute potential of 5 V at a distance of 1 meter, then:

$$k\cdot q = r\varphi = 1m\cdot 5V = 5\space V\cdot m$$

Then the potential due to 'q' at a distance of 3 meter will be:

$$\varphi = \dfrac{k\cdot q}{r} = \dfrac{5 V\cdot m}{3m} = 5/3V \approx 1.67V$$

Answer. 1.67 V.