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

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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 = 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 dV/dx = \frac{- qr(r - 2x)}{4$\pi$ \in_0 x^2 (r - x )^2} Therefore, x = r/2. It can be observed that dV/dx < 0 for x < r/2. Thus, the potential is decreasing first. At x = r/2, the potential is minimum. As dV/dx > 0 for x > r/2, the potential is increasing after x = r/2.

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