Q. An uniform electric field $E$ exists along positive $x$-axis. The work done in moving a charge 0.5 C through a distance 2 m along a direction making an angle $60^{o}$ with $x$-axis is 10 J. Then the magnitude of electric field is
A) $5\ Vm^{- 1}$
B) $2\ Vm^{- 1}$
C) $\sqrt{5}Vm^{- 1}$
D) $20\ Vm^{- 1}$
Explanation
(d) Force acting on the charged particle $= q\overrightarrow{E}$
Work done in moving a distance $S,$
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$W = q\overrightarrow{E}.\overrightarrow{S} = (qE) \times S \times \cos\theta$
$10J = (0.5\ C) \times E \times 2\cos 60^{o}$
$\Rightarrow E = 10 \times 2 = 20\ NC^{- 1} = 20\ Vm^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.