A parallel plate capacitor is formed by two p | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The maximum electric field $E$ that the dielectric can withstand is its dielectric strength,$E = 3.6 	imes 10^{7} \, Vm^{-1}$.The capacitance of

Vedclass · Accepted Answer

$2.33$

Vedclass · Answer

(D) The maximum electric field $E$ that the dielectric can withstand is its dielectric strength,$E = 3.6 	imes 10^{7} \, Vm^{-1}$.The capacitance of a parallel plate capacitor with a dielectric is $C = \frac{K \varepsilon_{0} A}{d}$.The maximum charge $q$ is given by $q = CV = C(Ed) = \left( \frac{K \varepsilon_{0} A}{d} ight) Ed = K \varepsilon_{0} A E$.Rearranging for the dielectric constant $K$,we get $K = \frac{q}{\varepsilon_{0} A E}$.Given $q = 7 	imes 10^{-6} \, C$,$A = 30 \pi 	imes 10^{-4} \, m^{2}$,$E = 3.6 	imes 10^{7} \, Vm^{-1}$,and $\frac{1}{4 \pi \varepsilon_{0}} = 9 	imes 10^{9} \, Nm^{2}C^{-2}$,which implies $\varepsilon_{0} = \frac{1}{36 \pi 	imes 10^{9}}$.Substituting these values: $K = \frac{7 	imes 10^{-6}}{\left( \frac{1}{36 \pi 	imes 10^{9}} ight) 	imes (30 \pi 	imes 10^{-4}) 	imes (3.6 	imes 10^{7})}$.$K = \frac{7 	imes 10^{-6} 	imes 36 \pi 	imes 10^{9}}{30 \pi 	imes 10^{-4} 	imes 3.6 	imes 10^{7}} = \frac{7 	imes 36 	imes 10^{3}}{30 	imes 3.6 	imes 10^{3}} = \frac{252}{108} = 2.33$.

Similar Questions

$A$ 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 $?$

$A$ capacitor is charged. When a dielectric slab of thickness $t = 4 \times 10^{-5} \ m$ is inserted between the plates,the distance between the plates has to be increased by $d' = 3.5 \times 10^{-5} \ m$ to maintain the same voltage. What is the dielectric constant $K$ of the dielectric?

Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$,then the potential difference across the other capacitor will become:

Two parallel metal plates having charges $+Q$ and $-Q$ face each other at a certain distance between them. If the plates are now dipped in a kerosene oil tank,the electric field between the plates will

Vedclass Test Series

Exam Paper Generator

Online Exam Module