The work done in placing the dielectric slab inside one of the capacitors as shown in diagram

Options

A. `(CV^(2))/(2) ((K-1)/(K+1))`
B. `(CV^(2))/(4)((K-1)/(k+1))`
C. `(CV^(2))/(4)((K+1)/(K-1))`
D. `(CV^(2))/(2)((K+1)/(k-1))`

Answer

B

Explanation

`U_(f) =1/2 ((CxxKC)/(C+KC)) V^(2) = (KCV^(2))/(2(1+K))`
`W = DeltaU =U_(f) -U_(i)`
`=(KCV^(2))/(2(1+K))=(CV^(2))/(4) =(CV^(2))/(2) [(k)/(1+k) -1/2]`
`=(CV^(2))/(2) [(2k-1-K)/(2(1+k))] = (CV^(2))/(2) ((K-1)/(K+1))`