A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then battery is disconnected. A slab of dielectric constant K is then inserted between the plates. If q, E and W denotes the magnitude of charge on each plate, electric field between the plates (After the slab insertion) and work done on the system, then
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a
q=ε0AVd
b
E=VKd
c
W=ε0AV22d1K-1
d
All of these
answer is D.
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Detailed Solution
C1=KC=Kε0Ad;V'=qC'=VKq= become same =ε0AVdE'=Vd=VKdW=Uf-Ui=q22C1K-1