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Hence obtain the work done in bringing a charge of $2 \times 10^{–9}\; C$ from infinity to the point P. Does the answer depend on the path along which the charge is brought?

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$q= 2 \times 10^{-9}$
Work done (W) =?
$W=qV= 2 \times 10^{-9} C \times 4 \times 10^{4}V$
$\quad= 8 \times 10^{-5} J$
No, work done will be path independent.
Any arbitrary infinitesimal path can be resolved into two perpendicular displacements:
One along r and another perpendicular to r. The work done corresponding to the later will be zero.
answered Jun 7, 2014 by meena.p

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