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In the network shown in figure, 1 and 2 are identical parallel plate capacitors, potential difference across capacitor 1 is 12 volt. Now a dielectric slab having k = 3 is inserted into the space between the plates of capacitor 1 and the slab completely fills the space. What will be the potential difference across capacitor 2?

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18 volt

20 volt

8 volt

16 volt

answer is B.

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Detailed Solution

E = emf of the battery = (12+12) volt = 24 volt.

After the slab is inserted in capacitor 1, equivalent capacitance

$C_{e} = \frac{kcxc}{kc + c} = \frac{3 c}{3 + 1} = \frac{3 c}{4}$

$∴$ Q = Charge on each capacitor $= \frac{3 c}{4} x 24 = 18 c$

$∴$ Potential difference between the plates of 2 $= \frac{18 c}{c} = 18 volt$

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