Two positive charges Q and 4Q 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 is minimum at P on the through A and B. What is (are) the distance(s) of P from A?

Options

A. `d/3` units to the right of A
B. `d/3` units to the left of A
C. `d/5` units to the right of A
D. `d` units to the left of A

Answer

A

Explanation

Let a unit positive charge be placed at point P between points A and B and at a distance x from A. ltbgt
Total potential at point P due to charge Q and 4Q,
`V = Q/x + (4Q)/((d-x))`
For the potential to be minimum at point P,
`(dV)/(dx) = 0, or, d/(dx)[Q/x + (4Q)/(d-x)] = 0, or, x = d/3`
`:.` The electric potential is minimum at a distance `d/3` from point A on its right.