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(C) Let the distance between points $A$ and $B$ be $d$. Let the charges at $A$ and $B$ be $+q$. At any point $P$ on the line segment $AB$ at a distanc

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decreases then increases

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(C) Let the distance between points $A$ and $B$ be $d$. Let the charges at $A$ and $B$ be $+q$. At any point $P$ on the line segment $AB$ at a distance $x$ from $A$, the distance from $B$ is $(d - x)$. The total electric potential $V$ at point $P$ is the sum of potentials due to both charges: $V = \frac{kq}{x} + \frac{kq}{d - x}$ To find the variation, we differentiate $V$ with respect to $x$: $\frac{dV}{dx} = -\frac{kq}{x^2} + \frac{kq}{(d - x)^2}$ Setting $\frac{dV}{dx} = 0$ for the minimum potential: $\frac{1}{x^2} = \frac{1}{(d - x)^2} \implies x = d - x \implies x = d/2$. At $x = d/2$, the potential is minimum. Thus, as we move from $A$ to $B$, the potential first decreases until the midpoint and then increases.

Two equal positive point charges are kept at points $A$ and $B$. What happens to the electric potential while moving from $A$ to $B$ along the straight line connecting them?

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