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:

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Solution

The correct option is C
Let C be the capacity without slab. Then after removing the slab, net capacity is C/2

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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}

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

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: