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Parth · Answer

See in this case Work done = change in potential energy + K.E. here KE = 0 assuming charge At a = 1 so U1 – U2 U1 intial Potential Energy U1 = k*(q^2)/a + 2*k*q*1/a for final configuration U2 = k*(q^2)/a + 2*k*(q)/(a/sqrt(3)) so |U1- U2| = (2*k*q*(3^(1/2) - 1))/a So option B pls Approve if it helps

Ayush_Deep · Answer

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 PE The initial potential energy is 2kq/a a is side of equilateral triangle. Final potential energy is 2kq/(a/3½)=2√3kq/a Work done = kq(2√3-2)/a= 2kq(√3-1)/a Option B . Thanks

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Sir answer for the question which I have attached as a image

Sir answer for the question which I have attached as a image

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