Question

Two equal positive charges are kept at points A and B. The electric potential at the points between A and B (excluding these points) is studied while moving from A to B. The potential

• A. continuously increases
• B. continuously decreases
• C. increases then decreases
• D. decreases then increases

Answer 1

The correct option is D decreases then increases

Let distance between A and B be d

Then potential at any distance $x$ from point A is given by

$V=\frac{kq}{x}+\frac{kq}{(d-x)}$

$V=kq(\frac{1}{x}+\frac{1}{(d-x)})$

Now

$dV/dx=\frac{kqd(2x-d)}{\left(x\right(d-x){)}^2}$

therefore for $x<d/2$

$dV/dx<0$ therefore potential decreases

and for $x>d/2$

$dV/dx>0$ therefore potential increases

Hence first decreases then increases option (D)