Questions
P is a point at a distance r1 from a point charge Q and potential at P is found to be 5 volt.. Now another point charge 4Q is placed at a distance r2 from P such that intensity of electric field at P becomes zero. Then final potential at P will be
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Correct option is D
Initially K.Qr1=V.Now, KQr 12=K.4Qr22⇒r2r1=4QQ=2.∴V|=KQr1+K.4Qr2=KQr11+4.r1r2=V1+4.12=3V ⇒V|=3×5 volt =15 volt.Similar Questions
A positive charge q and a negative charge -q are placed at x = -0 and x = + a respectively. The variation of V along X - axis is represented by the graph.
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