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 situated while moving from A to B. The potential  

पर्याय

• continuously increases 

•  continuously decreases

•  increases then decreases

• decreases then increases 

Answer 1

 decreases then increases

Let the distance between the points A and B be r.

Let us take a point P at a distance x from A (x < r). 

Electric potential V at point P due to two charges of equal magnitude q is given by

\[V = \frac{q}{4\pi \in_0 x} + \frac{q}{4\pi \in_0 (r - x)}\]
\[ \Rightarrow V = \frac{qr}{4\pi \in_0 x(r - x)}\]

Now, differentiating V with respect to x, we get

\[\frac{dV}{dx} = \frac{- qr(r - 2x)}{4\pi \in_0 x^2 (r - x )^2}\]

Therefore, x = r/2.
It can be observed that  \[\frac{dV}{dx} < 0\]  for x < r/2. 

Thus, the potential is decreasing first.

At  x = r/2, the potential is minimum.
As 

\[\frac{dV}{dx} > 0\]  for x > r/2, the potential is increasing after x = r/2.