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Standard XII

Physics

EMF and EMF Devices

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 constant K. $Q_{1}$ and $Q_{2}$ are the charges stored in $C_{1}$ and $C_{2}$ . Now, the dielectric slab is removed and the corresponding charges are $Q_{1}^{'}$ $Q_{2}^{'}$ . 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 C $\frac{Q_{2}^{'}}{Q_{2}} = \frac{K + 1}{2 K}$
Before the slab is removed, if $C_{1} = C, C_{2} = K C$
$C_{n e t} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{C K C}{C (1 + K)} = \frac{K C}{K + 1}$
$∴ Q_{1} = Q_{2} = \frac{K C E}{K + 1}$
After the slab is removed, $C_{1} = C_{2} = C$
$C_{n e t} = \frac{C}{2} ∴ Q_{1}^{'} = Q_{2}^{'} = \frac{C E}{2}$
$\frac{Q_{2}^{'}}{Q_{2}} = \frac{\frac{C E}{2}}{\frac{K C E}{(K + 1)}} = \frac{K + 1}{2 K}$

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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 connected in series to a battery as shown in figure. Capacitor 2 contains a dielectric slab of dielectric constant K as shown. If

Q_{1}

is the charge on each capacitor before removing the slab and

Q_{2}

is the charge on each capacitor after
moving the slab, then the correct relation between

Q_{1}

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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

C_{1}

and

C_{2}

are connected to a battery in series. A dielectric slab of dielectric constant

K = 2

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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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. Q1 andQ2 are the charges stored in C1 and C2. Now, the dielectric slab is removed and the corresponding charges are Q′1 Q′2. Then

The correct option is C Q′2Q2=K+12KBefore the slab is removed, if C1=C,C2=KC Cnet=C1C2C1+C2=CKCC(1+K)=KCK+1 ∴Q1=Q2=KCEK+1 After the slab is removed,C1=C2=C Cnet=C2∴Q′1=Q′2=CE2 Q′2Q2=CE2KCE(K+1)=K+12K