Question

Two charges are kept at a distance $d$. If the distance between them increases to four times, how does the electric potential between them change?

• A. It is doubled
• B. It is halved
• C. It is quartered
• D. It is quadrupled
• E. It is unchanged

Answer 1

Answer: (C)

It is quartered

Let the initial distance between the charges be $r$.

Thus initial electric potential $V=\frac{kq}{r}$

Now the distance between them is increased to 4 times i.e $r^{\prime }=4r$

$\therefore$ New electric potential $V^{\prime }=\frac{kq}{r^{\prime }}=\frac{kq}{4r}=\frac{V}{4}$

Thus electric potential is quartered.