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(Vf−Vi)=−qΔV ⇒ For movement of same charge (e−) from C→D and D→E, ΔV=5 Volt ∴WC→D=WD→E 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 e−will increase when it moves against→E.

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