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Byju's Answer

Standard XII

Physics

Displacement Vector

In the adjoin...

Question

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

\frac{K C E}{K + 1}

From

B

C

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\frac{K C E}{K + 1}

From

C

B

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\frac{(K - 1) C E}{2 (K + 1)}

From

B

C

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\frac{(K - 1) C E}{2 (K + 1)}

From

C

B

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Solution

The correct option is A $\frac{K C E}{K + 1}$ From $B$ to $C$
Here,

The capacitance of the capacitors are $C$

After dielectric inserted,

Capacitance are $C$ and $K C$

Capacitor are in series,

$C_{e q} = \frac{C \times K C}{K C + C} = \frac{K C}{K + 1}$

$Q = C_{e q} E = \frac{K C E}{K + 1}$

And the flow of charge is from cathode to anode( i.e - to + of battery).

Option $A$ is the correct answer

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