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TWO CONDENSER OF CAPACITIES 2C AND C ARE JOINED IN PARALLEL AND CHARGED UPTO POTENTIAL V .THE BATTERY IS REMOVED AND THE CONDENSER OF CAPACITY C IS FILLED COMPLETELY WITH A MEDIUM OF DIELECTRIC CONSTANT K . THE P.D ACROSS CAPACITOR WILL NOW BE.
1.3V/K+2
2.3V/K
3. V/K+2
4. V/K

iitkgp_bipin

Both the capacitors are charged to potential V.

Hence charges on them are CV and 2CV.

When capacitor C is filled with a medium of dielectric constant k then its capacitance value becomes kC.

Now let the charge on kC be q₁ and on 2C be q₂.

Since they are in parallel they have same potential difference.

q₁/kC = q₂/2C..................(1)

Now conserve charge i.e. initial charge = final charge

CV + 2CV = q₁ + q₂ .................(2)

From (1) and (2) :

q₂ = 6CV/(k+2)

Hence PD across capacitor = q₂/2C = 3V/(k+2)

1st option is correct.

rukrukkhan

SEE U AT 10:17

rukrukkhan

SEE U AT 10:17

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