Answer to Question #129778 in Physics for mah nooor

Question #129778

A point charge 'q' produce an absolute potential of 5 V at a distance of 1 meter. What will be the Absolute Potential due to 'q' at a distance of 3 meter ?


1
Expert's answer
2020-08-17T08:57:55-0400

By definition, the absolute potential of point charge at the distance rr is:


φ=kqr\varphi = k\dfrac{q}{r}

where kk is a constant.

A point charge 'q' produce an absolute potential of 5 V at a distance of 1 meter, then:


kq=rφ=1m5V=5 Vmk\cdot q = r\varphi = 1m\cdot 5V = 5\space V\cdot m

Then the potential due to 'q' at a distance of 3 meter will be:


φ=kqr=5Vm3m=5/3V1.67V\varphi = \dfrac{k\cdot q}{r} = \dfrac{5 V\cdot m}{3m} = 5/3V \approx 1.67V

Answer. 1.67 V.


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