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Question

A uniform electric field of magnitude 325 Vm−1 is directed in negative z direction. Coordinates of point A are (2 m,3 m) and those of point B are (4 m,5 m). Calculate the work done in bringing a unit positive charge along the given path.


A
2600 J
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B
3250 J
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C
5200 J
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D
zero
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Solution

The correct option is A 2600 J


Work done in moving the charge from C to B = 0
(since path is perpendicular to the electric field)

Work done in moving the charge from A to C
W=q(VfVi)=q(VCVA)=qΔV

Here, ΔV= Electric field × distance between A and C
=325V/m×8 m=2600 V

Hence, work done W=qΔV=2600 J

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