A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done on the system, in question, in the process of inserting the slab, then

Options

A. `Q = (epsilon_(0)AV)/(d)`
B. `Q = (epsilon_(0)AV^(2))/(2Kd)`
C. `E = (V)/(Kd)`
D. `W = (epsilon_(0)AV^(2))/(2d) (1 - (1)/(K))`

Answer

D

Explanation

After inserting the dielectric slab
New capacitance `C' = K.C = (K epsilon_(0)A)/(d)`
New potential difference `V' = (V)/(K)`
New charge `Q' = C'V' = (epsilon_(0)AV)/(d)`
New electric field `E' = (V')/(d) = (V)/(Kd)`
Work done `(W) =` Final energy - Initial energy
`W = (1)/(2)C'V'^(2) - (1)/(2)CV^(2) = (1)/(2)(KC) ((V)/(K))^(2) - (1)/(2)CV^(2)`
`= (1)/(2)CV^(2) ((1)/(K) -1) = - (1)/(2)CV^(2) (1-(1)/(K))`
`= - (epsilon_(0)AV^(2))/(2d) (1-(1)/(K))` so `|W| = (epsilon_(0)AV^(2))/(2d) (1-(1)/(K))`