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Question

A charge of 5 C experiences a force of 5000 N when it is kept in a uniform electric field. Find the potential difference (inV) between two points seperated by a distance of 1 cm.

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Solution

Given,

Charge, q=5 C
Force experienced by charge, F=5000 N
Seperation between given two points for which the potential difference is calculated is d=1 cm

When a charge is kept in an electric field E, Force acting on the charge is given by
F=qE
We know from the relation between uniform electric field E and potential V.
E=Vd

F=q(Vd)
Substituting the given data we get,
5000=5×V102

V=10 volt

Accepted answer: 10.

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