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प्रश्न
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
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उत्तर
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.
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