Q.

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

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a

10 V

b

250 V

c

1000 V

d

2500 V

answer is A.

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Detailed Solution

F=QE=QVd⇒5000=5×V10−2⇒V=10volt
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