Q 2 electrostatic

​Q2. Two isolated spherical shell having charges Q1 and Q2 and radii r1 and r2 are kept at very large separation. If they are joined by a conducting wire, then choose incorrect statement:

(A) Electrostatic potential energy of system must decrease.

(B) Electrostatic potential energy of system may decrease.

(C) If Q1r2 = Q2  r1, then no charge flow through the conducting wire.

(D) If Q1r2= Q2 r1, then electrostatic potential energy of system already minimum.

Dear Student,

potential of the sphere 1V=kQ1r1potential of the sphere 2V'=kQ2r2for no flow of charge V=V'kQ1r1=kQ2r2Q1r2=Q2r1  option C is correct .option D is also correct as the potential are equal already .let the charge transfer from sphere 1 to2 is qand the potential becomes equal kQ1-qr1=kQ2+qr2Q1r2-qr2=Q2r1+qr1qr1+r2=Q2r1-Q1r2q=Q2r1-Q1r2r1+r2hence the electrostatic PE will decreases i any case and will be minimum when V=V'option B is correct it may decrease Regards

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