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The work done in carrying a charge of 5 $μ$ C from point A to a point B in an electric field is 10 mJ. The potential difference (V_B- V_A) is then

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+2 kV

+200 V

-200 V

-20 kV

answer is A.

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Detailed Solution

$W = q (V_{A} - V_{B})$
$∴$ potential difference $= \frac{W}{q} = \frac{10 mJ}{5 μC} = 2 kV$

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