NEET-XII-Physics
29: Electric Field and Potential
- #4Two 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
(a) continuously increases
(b) continuously decreases
(c) increases then decreases
(d) decreases then increasesdigAnsr: dAns : (d) 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{\,\mathrm{\,d\,}V}{\,\mathrm{\,d\,}x}=\frac{-qr(r-2x)}{4\pi {\in }_{0}{x}^{2}(r-x{)}^{2}}``
Therefore, x = r/2.
It can be observed that `` \frac{\,\mathrm{\,d\,}V}{\,\mathrm{\,d\,}x}<0`` for x < r/2. Thus, the potential is decreasing first. At
x = r/2, the potential is minimum.
As `` \frac{\,\mathrm{\,d\,}V}{\,\mathrm{\,d\,}x}>0`` for x > r/2, the potential is increasing after x = r/2.
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