In the diagram,the area of each plate is A = | vedclass

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

Question

(C) The system consists of two capacitors in parallel connected to plate '$Q$'.One capacitor is formed by plates '$P$' and '$Q$' with separation $d$,a

Vedclass · Accepted Answer

$\frac{20}{3} \,V$

Vedclass · Answer

(C) The system consists of two capacitors in parallel connected to plate '$Q$'.One capacitor is formed by plates '$P$' and '$Q$' with separation $d$,and the other by plates '$Q$' and '$R$' with separation $2d$.The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.Let $C_1$ be the capacitance between '$P$' and '$Q$': $C_1 = \frac{\epsilon_0 A}{d}$.Let $C_2$ be the capacitance between '$Q$' and '$R$': $C_2 = \frac{\epsilon_0 A}{2d} = \frac{C_1}{2}$.Since both plates '$P$' and '$R$' are grounded,the potential of both is $0 \,V$. Thus,the capacitors are in parallel with respect to plate '$Q$'.The equivalent capacitance is $C_{eq} = C_1 + C_2 = C_1 + \frac{C_1}{2} = \frac{3}{2} C_1$.Substituting the values: $C_1 = \frac{8.85 	imes 10^{-12} 	imes 2}{2 	imes 10^{-3}} = 8.85 	imes 10^{-9} \,F$.$C_{eq} = \frac{3}{2} 	imes 8.85 	imes 10^{-9} = 1.3275 	imes 10^{-8} \,F$.The potential of plate '$Q$' is $V = \frac{q}{C_{eq}} = \frac{8.85 	imes 10^{-8}}{1.3275 	imes 10^{-8}} = \frac{8.85}{1.3275} = \frac{8.85}{1.5 	imes 8.85} 	imes 10 = \frac{20}{3} \,V$.

In the diagram,the area of each plate is $A = 2 \,m^2$ and $d = 2 \times 10^{-3} \,m$. An electric charge of $q = 8.85 \times 10^{-8} \,C$ is given to the plate '$Q$'. Plates '$P$' and '$R$' are connected to the ground. Then the potential of plate '$Q$' is:

Similar Questions

Assertion : The total charge stored in a capacitor is zero.
Reason : The field just outside the capacitor is $\frac{\sigma }{{{\varepsilon _0}}}$. ( $\sigma $ is the charge density).

$A$ capacitor is made of a flat plate of area $A$ and a second plate of stair-like structure as shown in the figure. The area of each stair is $\frac{A}{3}$ and the height is $d$. The capacitance of the arrangement is

The capacity of a parallel plate capacitor depends on:

The plates of a parallel plate capacitor are pulled apart with a velocity $v$. If at any instant their mutual distance of separation is $d$,then the magnitude of the time rate of change of capacity depends on $d$ as follows:

Five identical capacitor plates,each of area $A$,are arranged such that adjacent plates are at a distance $d$ apart. The plates are connected to a source of $emf$ $V$ as shown in the figure. Then the charges on plates $1$ and $4$ are,respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module