Two equal charges q of opposite sign are separated by a small distance '2l'. The electric potential at a point on the perpendicular bisector of the line joining the two charges at a distance r is :
A
14πε0qr
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B
14πε02qr
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C
Zero
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D
14πε02qr2
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Solution
The correct option is A Zero V=Kq√r2+l2+K(−q)√r2+l2