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Question

Two identical parallel plate capacitors are connected in series to a battery of 100 V. A dielectric slab of dielectric constant 4 is inserted between the plates of the second capacitor to fill the space between its plates, completely. The potential difference across the capacitors will now be, respectively.

A
50 V, 50 V
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B
80 V, 20 V
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C
20 V, 80 V
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D
75 V, 25 V
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Solution

The correct option is B 80 V, 20 V
Let C be the initial capacitance of both capacitor.

And we know that when dielectric is inserted between the plates, the capacitance increases by K times. So, the capacitance of second capacitor will be

C=KC=4C


So, their equivalent capacitance will be,

Ceq=C×4CC+4C=4C5

In series combination, charge on each capacitor remains the same, so the charge on each capacitor,

Q=CeqV=4C5×100=80C

Now, Voltage across C is,

VC=QC=80CC=80 V

And similarly, voltage across 4C is

V4C=QC=80C4C=20 V

Hence, option (B) is correct.
Why this Question?
Concept:When dielectric is inserted between the plates of the capacitor, its capacitance increases to initial capacitance times dielectric constant.

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