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Physics Question on Electrostatic potential

The work done on a particle of mass mm by a force K[x(x2+y2)3/2i^+y(x2+y2)3/2j^]K \left[\frac{ x }{\left( x ^{2}+ y ^{2}\right)^{3 / 2}} \hat{ i }+\frac{ y }{\left( x ^{2}+ y ^{2}\right)^{3 / 2}} \hat{ j }\right] ( KK being a constant of appropriate dimensions, when the particle is taken from the point (a,0)( a , 0) to the point (0,a)(0, a ) along a circular path of radius a about the origin in the xx-y plane is

A

2Kπa\frac{ 2K \pi}{a}

B

Kπa\frac{ K \pi}{a}

C

Kπ2a\frac{ K \pi}{2a}

D

0

Answer

0

Explanation

Solution

dw=Fdr=F(dxi^+dj^)d w =\vec{ F } \cdot d \vec{ r }=\vec{ F } \cdot( d x \hat{ i }+ d \hat{ j })
=Kxdx(x2+y2)3/2+ydy(x2+y2)3/2= K \int \frac{ x dx }{\left( x ^{2}+ y ^{2}\right)^{3 / 2}}+\frac{ ydy }{\left( x ^{2}+ y ^{2}\right)^{3 / 2}}
x2+y2=a2x ^{2}+ y ^{2}= a ^{2}
w=Ka3a0xdx+0aydyw =\frac{ K }{ a ^{3}} \int\limits_{ a }^{0} x d x +\int\limits_{0}^{ a } ydy
=Ka3(a22+a22)=0=\frac{ K }{ a ^{3}}\left(\frac{- a ^{2}}{2}+\frac{ a ^{2}}{2}\right)=0