A parallel plate capacitor with plate area A | vedclass

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

Question

(B) Consider an element of width $dx$ at a distance $x$ $(x < d/2)$ from the left plate. The differential capacitance $dC$ of this element is given by

Vedclass · Accepted Answer

$\frac{kA}{2 \ln \left(\frac{2\varepsilon_{0} + kd}{2\varepsilon_{0}}\right)}$

Vedclass · Answer

(B) Consider an element of width $dx$ at a distance $x$ $(x For the first half of the capacitor ($0$ to $d/2$),the capacitors are in series. The equivalent capacitance $C_1$ is given by:$\frac{1}{C_1} = \int_{0}^{d/2} \frac{1}{dC} = \frac{1}{A} \int_{0}^{d/2} \frac{dx}{\varepsilon_{0} + kx}$Integrating this,we get:$\frac{1}{C_1} = \frac{1}{kA} [\ln(\varepsilon_{0} + kx)]_{0}^{d/2} = \frac{1}{kA} \ln\left(\frac{\varepsilon_{0} + kd/2}{\varepsilon_{0}}\right) = \frac{1}{kA} \ln\left(\frac{2\varepsilon_{0} + kd}{2\varepsilon_{0}}\right)$So,$C_1 = \frac{kA}{\ln\left(\frac{2\varepsilon_{0} + kd}{2\varepsilon_{0}}\right)}$.Due to symmetry,the capacitance of the second half $(C_2)$ is the same as $C_1$. Since the two halves are in series,the total equivalent capacitance $C_{eq}$ is:$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{C_1}{2} = \frac{kA}{2 \ln \left(\frac{2\varepsilon_{0} + kd}{2\varepsilon_{0}}\right)}$.

Similar Questions

$A$ capacitor is half-filled with a dielectric $(K=2)$ as shown in Figure $A$. If the same capacitor is to be filled with the same dielectric as shown in Figure $B$, what would be the thickness $t$ of the dielectric so that the capacitor still has the same capacity?

Consider a parallel plate capacitor of $10\,\mu F$ with air filled in the gap between the plates. Now,one half of the space between the plates is filled with a dielectric of dielectric constant $K = 4$,as shown in the figure. The capacity of the capacitor changes to.......$\mu F$.

$A$ parallel plate air capacitor has capacity $C$ farad, potential $V$ volt, and energy $E$ joule. When the gap between the plates is completely filled with a dielectric of dielectric constant $K$, what happens to the potential $V$ and energy $E$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module