A particle with a charge of 3 coulombs is taken from a point at a potential of 50 V to another point at a potential of 120 V. Calculate the work done.

23.33 J

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2.4 J

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210 J

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40 J

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Solution

The correct option is C 210 J
We know that the work done is the change in potential energy.
So, $W = U_{f} - U_{i} = q V_{f} - q V_{i} = q (V_{f} - V_{i}) = 3 (120 - 50) = 210 J$

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