A capacitor is connected to a battery of voltage V . Now a di electric slab of dielectric constant k

English

Gujarati

Hindi

2. Electric Potential and Capacitance

medium

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 $q_{0}$ )

A

$\frac{\varepsilon_{0} A}{d} V$

B

$\frac{k \varepsilon_{0} A}{d} V$

C

$\frac{\varepsilon_{0} A}{k d} V$

D

zero

(AIIMS-2019)

Solution

The initial charge on the capacitor is,

$q_{0}=C_{0} V$

$=\frac{\varepsilon_{0} A}{d} V$

The charge on the capacitor when di$-$electric is inserted is,

$q=k q_{0}$

$=\frac{k \varepsilon_{0} A}{d} V$

Standard 12

Physics

Share

Similar Questions

An insulator plate is passed between the plates of a capacitor. Then the displacement current

medium

The capacitance of a parallel plate capacitor with air as medium is $6\, \mu F$. With the introduction of a dielectric medium, the capacitance becomes $30\, \mu F$. The permittivity of the medium is……….$C ^{2} N ^{-1} m ^{-2}$

$\left(\varepsilon_{0}=8.85 \times 10^{-12} C ^{2} N ^{-1} m ^{-2}\right)$

easy

(NEET-2020)

Adielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?

medium

Define dielectric constant.

medium

In a parallel plate condenser, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. There is a glass slab of $3\,mm$ thick and of radius $12\,cm$ with dielectric constant $6$ between its plates. The capacity of the condenser will be

medium