A parallel plate capacitor is to be designed, using a dielectric of dielectric constant 5 , so as to

English

Gujarati

Hindi

2. Electric Potential and Capacitance

medium

A parallel plate capacitor is to be designed, using a dielectric of dielectric constant $5$, so as to have a dielectric strength of $10^9\;Vm^{-1}$ . If the voltage rating of the capacitor is $12\;kV$, the minimum area of each plate required to have a capacitance of $80\;pF$ is

A

$10.5 \times 10^{-6} \;m ^2$

B

$25.0 \times 10^{-5}\; m ^2$

C

$12.5 \times 10^{-5}\; m ^2$

D

$21.7 \times 10^{-6} \;m ^2$

(NEET-2017)

Solution

$V=E d$

$\therefore d=\frac{V}{E}$

$C=\frac{K \varepsilon_0 A}{d}$

$\therefore A=\frac{C d}{K \varepsilon_0}$

$=\frac{C V}{K E \varepsilon_0}$

$=\frac{80 \times 10^{-12} \times 12 \times 10^3}{5 \times 10^9 \times 8.85 \times 10^{-12}}$

$=21.69 \times 10^{-6}$

$=21.7 \times 10^{-6} \;m ^2$

Standard 12

Physics

Share

Similar Questions

Separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t(t < d)$ is introduced between the plates, its capacitance becomes

hard

A parallel plate air capacitor of capacitance $C$ is connected to a cell of emf $V$ and then disconnected from it. A dielectric slab of dielectric constant $K,$ which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect $?$

easy

(AIPMT-2015)

A parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is

easy

A parallel-plate capacitor of area $A,$ plate separation $d$ and capacitance $C$ is filled with four dielectric materials having dielectric constants $K_1,K_2,K_3$ and $K_4$ as shown in the figure. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $K$ is given by

medium

(NEET-2016)

The electric field between the plates of a parallel plate capacitor when connected to a certain battery is ${E_0}$. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant $K$ without disturbing the battery connections, the field between the plates shall be

easy