First slide

Potential energy of single charge in external field

Question

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

Easy

A

+2 KV.
B

-2 KV
C

+ 200 V
D

-200 V

Solution

$w_{AB} = 5 μC (V_{A} - V_{B}) \Rightarrow \frac{10 mJ}{5 μC} = V_{A} - V_{B} \begin{aligned} ∴ V_{A} - V_{B} = 2 kV \\ or V_{B} - V_{A} = - 2 kV \end{aligned}$

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