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Question

The figure shows a set of equipotential surfaces. There are a few points marked on them. An electron is being moved from one point to other. Which of the following statements is incorrect?


A
The electric field is directed along +x axis.
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B
Work done by the electric field, in moving the electron from B to C, is positive.
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C
Work done by the electric field, in moving the electron from C to D is the same as from D to E.
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D
As the electron moves from E to A, the potential energy increases
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Solution

The correct option is A The electric field is directed along +x axis.
Electric field lines are always directed from region of high potential to low potential. Besides the E is perpendicular to the equipotential surfaces.


E is along ve x axis

option (a) is incorrect.

Electron being a ve charged particle, it will experience force in a direction opposite to E. i.e. along +ve x axis.

Thus, F and displacement (s) are in the same direction when electron displaced from B to C.

WElectric force+ve

(b) is correct.

Work done by electric force is, W=q(VfVi)=qΔV

For movement of same charge (e) from CD and DE, ΔV=5 Volt

WCD=WDE

Also, electron is moving against the electric field hence its P.E will increase option (c) & (d) are correct.


Why this question?Whenever work is done against a conservativefield, the associated P.E of body will increaseExample: on throwing a ball upwards, it is movingagainst gravitational field. Hence, its gravitationalP.E increases. Similarly electrostatic P.E of ewill increase when it moves againstE.

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