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Byju's Answer

Standard XII

Physics

New Capacitance in Dielectric

Two identical...

Question

Two identical capacitors $1$ and $2$ are connected in series to a battery as shown in figure. Capacitor $2$ contains a dielectric slab of dielectric constant $k$ as shown. $Q_{1}$ and $Q_{2}$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $Q_{1}^{'}$ and $Q_{2}^{'}$ .
Then

\frac{Q_{1}^{'}}{Q_{1}} = \frac{k + 1}{k}

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\frac{Q_{2}^{'}}{Q_{2}} = \frac{k + 1}{2}

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\frac{Q_{2}^{'}}{Q_{2}} = \frac{k + 1}{2 k}

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\frac{Q_{1}^{'}}{Q_{1}} = \frac{k}{2}

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Solution

The correct option is B $\frac{Q_{2}^{'}}{Q_{2}} = \frac{k + 1}{2 k}$
The capacitor 1 has capacitance $C$ and as dielectric contains in capacitor 2 so its capacitance becomes $k C$ .
The net capacitance $C_{e q} = \frac{C . k C}{C + k C} = \frac{C k}{1 + k}$
and $Q_{e q} = C_{e q} E$
When dielectric is reemoved from capacitor 2, its capacitance becomes $C$ .
now net capacitance $C_{e q}^{'} = \frac{C . C}{C + C} = \frac{C}{2}$
and $Q_{e q}^{'} = C_{e q}^{'} E = \frac{C E}{2}$
When the capacitors are connected in series, they each have the same charge as the net capacitance. Thus, $Q_{e q} = Q_{1} = Q_{2}$ and $Q_{e q}^{'} = Q_{1}^{'} = Q_{2}^{'}$
$∴ \frac{Q_{1}^{'}}{Q_{1}} = \frac{Q_{2}^{'}}{Q_{2}} = \frac{2 k}{k + 1}$

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Similar questions

Two identical capacitors 1 and 2 are connected in series to a battery as shown in the figure. Capacitor 2 contains a dielectric slab of dielectric constant slab of dielectric constant K as shown. Q₁ and Q₂ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are then:

Q. Two identical capacitors 1 and 2 are connected in series to a battery as shown in figure. Capacitor 2 contains a dielectric slab of constant K.

Q_{1}

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Q. Two identical capacitors 1 and 2 connected in series to a battery as shown in figure. Capacitor 2 contains a dielectric slab of dielectric constant K as shown. If

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is the charge on each capacitor before removing the slab and

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Two identical capacitors 1 and 2 are connected in series. The capacitor 2 contains a dielectric slab of constant $K$ as show. They are connected to as battery of emf $V_{0}$ volts. The dielectric slab is then removed. Let $Q_{1}$ and $Q_{2}$ be the charge stored in the capacitors before removing the slab and ${Q_{1}}^{'},$ and ${Q_{2}}^{'}$ be the values after removing the slab. Then:

Q. Two identical sized capacitors

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is filled in the half region of both the capacitors as shown in figure.

(C \to

capacitance,

q \to

charge stored,

U \to

energy stored

)

Match the following column:

Column I	Column II
$(A)$ $\frac{C_{1}}{C_{2}}$	$(p)$ $8 / 9$ $(q)$ $4 / 9$ $(r)$ $9 / 8$ $(s)$ None of these
$(B)$ $\frac{q_{1}}{q_{2}}$
$(C)$ $\frac{U_{1}}{U_{2}}$

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