Question

1)When a charged particle moves from a higher equipotential surface to a lower equipotential surface what is the nature of the work done by the electric field? a) The work is negative b) the work is positive c) not enough information given d) the work is zero

2)When considering a point in the E-field of a single charge, the electric potential value at r=infinity, according to the common convention, is zero. All other values are a) positive b) zero c) negative d) It depends upon the sigh of the charge involved.

3)An electric dipole is created on the x axis of a coordinate system. The positive charge is at x = -0.2 m and the negative charge is at x = +0.2 m. Each charge has a magnitude of 2 nC. If point P is at x = 0.5 m and point Q is at x = 0.1 m, find Vpq. a) 7.5 volts b)10.3 volts c)34.3 volts d)85.7 volts

Answer 1

1.

W = q ( V2 - V1)

so workdone is also dependent on the type of charge either positive or negative, which isn't given any where  

option (c) is correct.

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2.

V = k q / r

option (d) is correct.

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3.

Vp = k ( 2 / 0.7 - 2/0.3) *10^-9 = - 34.3V

Vq = k ( 2 / 0.3 - 2 / 0.1)*10^-9 = - 120

Vp - Vq = - 34.3 - ( - 120) = 85.7 V

option (d) is correct

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do comment in case any doubt, will reply for sure.. Goodluck