A composite parallel plate capacitor is made up of two different dielectric materials with different

English

Gujarati

Hindi

2. Electric Potential and Capacitance

hard

A composite parallel plate capacitor is made up of two different dielectric materials with different thickness $\left(t_{1}\right.$ and $\left.t_{2}\right)$ as shown in figure. The two different dielectric material are separated by a conducting foil $F$. The voltage of the conducting foil is $.....V$

A

$6$

B

$66$

C

$600$

D

$60$

(JEE MAIN-2022)

Solution

Capacitance of each capacitor

$C _{1}=\frac{ A 3 \epsilon_{0}}{\frac{1}{2}}=6 A \epsilon_{0}$

$C _{2}= A 4 E _{0}=4 A \in_{0}$

Equivalent capacitance

$C _{\text {eq }}=\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}} \Rightarrow \frac{24}{10} A \in_{0}$

$q _{\text {net }}= C _{\text {eq }}(\Delta V ) \Rightarrow 240 A \in_{0}$

$\Delta V _{2}=\frac{240 A \epsilon_{0}}{4 A \epsilon_{0}}=60\,V$

$\left(\Delta V _{2}=\text { Potential drop across } C _{2}\right)$

$V _{\text {foil }}=60\,V$

Standard 12

Physics

Share

Similar Questions

Two identical capacitors $1$ and $2$ are connected in series. The capacitor $2$ contains a dielectric slab of constant $K$ as shown. They are connected to a 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

hard

The respective radii of the two spheres of a spherical condenser are $12\;cm$ and $9\;cm$. The dielectric constant of the medium between them is $ 6$. The capacity of the condenser will be

medium

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

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$, where $'d'$ is the separation between the plates of parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $\left( C _{0}\right)$ is given by the following relation

medium

(JEE MAIN-2021)

What is polarisation ?

easy