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Sir answer for the question which I have attached as a image
See in this case Work done = change in potential energy + K.E. here KE = 0assuming charge At a = 1so U1 – U2U1 intial Potential Energy U1 = k*(q^2)/a + 2*k*q*1/afor final configurationU2 = k*(q^2)/a + 2*k*(q)/(a/sqrt(3))so |U1- U2| = (2*k*q*(3^(1/2) - 1))/aSo option Bpls Approve if it helps
The correct answer will be option B .Work done in moving a charge particle is equal to change in the potential energy of particle = final PE - initial PEThe initial potential energy is 2kq/aa is side of equilateral triangle.Final potential energy is 2kq/(a/3½)=2√3kq/aWork done = kq(2√3-2)/a= 2kq(√3-1)/aOption B .Thanks
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