First slide

Potential energy

Question

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

Easy

A

+2 kV
B

-2 kV
C

+200 V
D

-200 V

Solution

$Work done W = Q (V_{B} - V_{A}) \Rightarrow (V_{B} - V_{A}) = \frac{W}{Q} = \frac{10 \times 10^{- 3}}{5 \times 10^{- 6}} J / C = 2 kV$

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