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Q.

The work done in carrying a charge of 5 μC from point A to a point B in an electric field is 10 mJ. The potential difference (VB- VA) is then

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a

+2 kV

b

+200 V

c

-200 V

d

-20 kV

answer is A.

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Detailed Solution

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W=qVA−VB
∴potential difference =Wq=10 mJ5 μC=2kV

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