In adjoining figure, capacitor (1) and (2) have a capacitance ‘C’ each. When dielectric

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2. Electric Potential and Capacitance

normal

In the adjoining figure, capacitor $(1)$ and $(2)$ have a capacitance $‘C’$ each. When the dielectric of dielectric consatnt $K$ is inserted between the plates of one of the capacitor, the total charge flowing through battery is

$\frac{{KCE}}{{K + 1}}$ from $B $ to $C$

$\frac{{KCE}}{{K + 1}}$ from $C$ to $B$

$\frac{{(K - 1)CE}}{{2(K + 1)}}$ from $B$ to $C$

$\frac{{(K - 1)CE}}{{2(K + 1)}}$ from $C$ to $B$

Solution

Before insert dielectric, the equivalent capacitance $C_{e q}=\frac{C \times C}{C+C}=C / 2$

$Q_{e q}=C_{e q} E=C E / 2$

As battery is connected to capacitor so $E$ remains constant.

After insert dielectric $, C_{e q}^{\prime}=\frac{K C C}{K C+C}=\frac{K}{K+1} C$

$Q_{e q}^{\prime}=C_{e q}^{\prime} E=\frac{K}{K+1} C E$

Charge flow through the battery is $\Delta Q=Q_{e q}^{\prime}-Q_{e q}=\frac{K}{K+1} C E-\frac{C E}{2}=\frac{(K-1) C E}{2(K+1)}$

As $\mathrm{E}$ is constant so for inserting dielectric the charge on the capacitor will increase and it will flow from capacitance to battery to maintain constant $E$

Standard 12

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hard

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(JEE MAIN-2018)

In the adjoining figure, capacitor $(1)$ and $(2)$ have a capacitance $‘C’$ each. When the dielectric of dielectric consatnt $K$ is inserted between the plates of one of the capacitor, the total charge flowing through battery is

Solution

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