An isolated capacitor of capacitance C is charged to a potential V. Then a dielectric slab of dielectric constant K is inserted as shown. The net charge on four surfaces 1,2, 3 and 4 would be respectively.

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$CV, \frac{- CV}{K}, \frac{CV}{K}, - CV$

$0, \frac{CV}{K}, \frac{- CV}{K}, 0$

0, CC, –CV, 0

CV, 0, 0, –CV

answer is B.

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Detailed Solution

Due to polarization, charge on dielectric slab would be $CV (1 - \frac{1}{K})$ .

So, Charge on 1 = 0

Charge on $2 = CV - CV (1 - \frac{1}{K}) = \frac{CV}{K}$

Charge on $3 = CV (1 - \frac{1}{K}) - CV = - \frac{CV}{K}$

Charge on 4 = 0

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