A parallel plate capacitor with air as medium between plates has a capacitance of 10,μ F . area of

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2. Electric Potential and Capacitance

medium

A parallel plate capacitor with air as medium between the plates has a capacitance of $10\,\mu F$. The area of capacitor is divided into two equal halves and filled with two media as shown in the figure having dielectric constant ${k_1} = 2$and ${k_2} = 4$. The capacitance of the system will now be.......$\mu F$

$10$

$20$

$30$

$40$

Solution

(c) ${C_R} = {C_1} + {C_2} = \frac{{{k_1}{\varepsilon _0}{A_1}}}{d} + \frac{{{k_2}{\varepsilon _0}{A_2}}}{d}$
$ = \frac{{2 \times {\varepsilon _0}\frac{A}{2}}}{d} + \frac{{4 \times {\varepsilon _0}\frac{A}{2}}}{d}$$ = 2 \times \frac{{10}}{2} + 4 \times \frac{{10}}{2} = 30\,\mu F$

Standard 12

Physics

The two metallic plates of radius $r$ are placed at a distance $d$ apart and its capacity is $C$. If a plate of radius $r/2$ and thickness $d$ of dielectric constant $6$ is placed between the plates of the condenser, then its capacity will be

hard

A parallel plate capacitor, partially filled with a dielectric slab of dielectric constant $K$ , is connected with a cell of emf $V\ volt$ , as shown in the figure. Separation between the plates is $D$ . Then

medium

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

easy

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

normal

A capacitor is connected to a battery of voltage $V$. Now a di electric slab of dielectric constant $k$ is completely inserted between the plates, then the final charge on the capacitor will be

(If initial charge is $q_{0}$ )

medium

(AIIMS-2019)

Solution

Similar Questions

The two metallic plates of radius $r$ are placed at a distance $d$ apart and its capacity is $C$. If a plate of radius $r/2$ and thickness $d$ of dielectric constant $6$ is placed between the plates of the condenser, then its capacity will be

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A capacitor is connected to a battery of voltage $V$. Now a di electric slab of dielectric constant $k$ is completely inserted between the plates, then the final charge on the capacitor will be

(If initial charge is $q_{0}$ )