First slide

Electrostatic potential energy

Question

There is an electric field E in X-direction. If the work done on moving a charge 0.2 C through a distance of 2 m along a line making an angle $60^{\circ}$ with the X-axis is 4.0, what is the value of E

Easy

A

$\sqrt{3} N / C$
B

4 N/C
C

5 N/C
D

20 $N C^{- 1}$

Solution

$W = qV = qE \cdot d$
$\Rightarrow 4 = 0 .2 \times E \times (2 \cos 60^{\circ})$
$= 0 .2 E \times (2 \times 0 .5)$
$∴ E = \frac{4}{0 .2} = 20 {NC}^{- 1}$

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