First slide

Capacitance

Question

Two identical parallel plate capacitors are connected in series and a 12 volt battery is connected across this combination. Now a dielectric slab having dielectric constant 2 is inserted into the space between the plates of one of the capacitors and the slab completely fills the space between the plates. Then potential difference between the plates of this capacitors is

Moderate

A

4.5 volt
B

6 volt
C

8 volt
D

4 volt

Solution

After the slab is inserted, final equivalent capacitance of the system,
$C_{e} = \frac{K C \times C}{K C + C} = \frac{K C}{K + 1} = \frac{2 C}{2 + 1} = \frac{2 C}{3}$
$∴$ Q = New charge in each capacitor = $C_{e} \times e = \frac{2 C}{3} \times 12 = 8 C$
$∴$ V= New potential difference between the plates
$= \frac{Q}{C_{B}} = \frac{8 C}{2 C} = 4 volt$

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