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If a dielectric slab is inserted in between a capacitor of capacitance C after charging it to a potential V, the net charge on surfaces shown below will be :

$(A)\;CV, \frac{-CV}{K} , \frac{CV}{K}, -CV \\ (B)\;0,CV,-CV,0\\ (C)\;0, \frac{CV}{K} , \frac{-CV}{K}, 0 \\ (D)\;none $

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Due to polarization charge on slab will be $CV (1- \large\frac{1}{K})$
Charge on $1=0$
Charge on $2= CV-CV (1-\large\frac{1}{k})$
$\qquad= \large\frac{CV}{k}$
Charge on $3=CV (1-\large\frac{1}{k})$$-CV$
$\qquad= \large\frac{-CV}{K}$
Charge on $4=0$
Hence c is the correct answer.


answered Jan 10, 2014 by meena.p
edited Aug 1, 2014 by thagee.vedartham

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