The plates of a parallel plate capacitor are | vedclass

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

Question

(B) The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$,where $\varepsilon_0$ is the permittivity of free sp

Vedclass · Accepted Answer

$1/d^2$

Vedclass · Answer

(B) The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$,where $\varepsilon_0$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the separation distance.To find the rate of change of capacitance with respect to time $t$,we differentiate $C$ with respect to $t$:$\frac{dC}{dt} = \frac{d}{dt} \left( \frac{\varepsilon_0 A}{d} \right) = \varepsilon_0 A \frac{d}{dt} (d^{-1})$.Using the chain rule,$\frac{dC}{dt} = \varepsilon_0 A (-d^{-2}) \frac{dd}{dt}$.Since the plates are pulled apart with velocity $v$,the rate of change of separation is $\frac{dd}{dt} = v$.Substituting this,we get $\frac{dC}{dt} = -\frac{\varepsilon_0 A}{d^2} v$.The magnitude of the rate of change is $\left| \frac{dC}{dt} \right| = \frac{\varepsilon_0 A v}{d^2}$.Thus,$\left| \frac{dC}{dt} \right| \propto \frac{1}{d^2}$.

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:

Similar Questions

Five identical plates each of area $A$ are joined as shown in the figure. The distance between the plates is $d$. The plates are connected to a potential difference of $V \text{ volts}$. The charge on plates $1$ and $4$ will be

The capacitance of a parallel plate capacitor does not depend on:

$A$ parallel plate capacitor has rectangular plates of area $400 \, cm^2$ separated by a distance of $2 \, mm$ with air as the medium. If a potential difference of $200 \, V$ is applied across the capacitor,what is the charge on the plates?

The charges on the four surfaces of two isolated plates with capacitance $C$ are $Q_1$,$Q_2$,$Q_3$,and $Q_4$ as shown in the figure. The potential difference between the plates is:

The force between the plates of a parallel plate capacitor of capacitance $C$ and distance of separation of the plates $d$ with a potential difference $V$ between the plates is

Vedclass Test Series

Exam Paper Generator

Online Exam Module