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.

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.

Toppr Admin · Accepted Answer

After inserting a dielectric plate C1-3CandC2-CV1-C-xD7-V3C-C-C-xD7-1004C-25V-V2-100-x2212-25-75V

After inserting a dielectric plate C1=3CandC2=CV1=C×V3C+C=C×1004C=25V,V2=100−25=75V

After inserting a dielectric plate $C_{1} = 3 C a n d C_{2} = C$
$V_{1} = \frac{C \times V}{3 C + C} = \frac{C \times 100}{4 C} = 25 V, V_{2} = 100 - 25 = 75 V$