Question

Two positive charges Q and $4$Q are placed at points A and B respectively, where B is at a distance 'd' units to the right of A. The total electric potential due to these charges in minimum at P on the line through A and B, What is (are) the distance(s) of P from A?

• A. $\frac{d}{3}$ units to the right of A
• B. $\frac{d}{3}$ units to the left of A
• C. $\frac{d}{5}$ units to the right of A
• D. d units to the left of A

Answer 1

The correct option is A $\frac{d}{3}$ units to the right of A

Electric potential is minimum where $\overrightarrow{E}=0$
here
$\frac{kQ}{x^2}=\frac{k4Q}{(d-x{)}^2}$
$\frac{1}{x}=\frac{2}{d-x}$
$d-x=2x$
$x=\frac{d}{3}$
$\frac{d}{3}$ units right of A.