Q.

A parallel plate capacitor of  plate area A and plate separation d is charged to potential V and then the battery is disconnected . A slab of dielectric constant K is then inserted between the plates of the capacitors 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 inserted) and work done on the system in question in the process of inserting the slab , then state incorrect from the following:

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a

Q=ε0AVd

b

W=ε0AV22kd

c

E=Vkd

d

W=ε0AV22d1−1k

answer is B.

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Detailed Solution

After inserting the dielectric slabNew capacitance  C'=K.C=Kε0AdNew potential difference  V'=VKNew charge  Q'=C'V'=ε0AVd   Charge remains same.New electric filed  E'=V'd=VKdWork done (W)= Final energy – initial energyW=12C'V'2−12CV2=12KCVK2−12CV2 =12CV21K−1=−12CV21−1K =−ε0AV22d1−1K ⇒W=ε0AV22d1−1K
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