Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Explain the topic cyclindrical capApacitor

Answers (1)

Cylinderical capacitor 

It consists of two coaxial cylinders of radii a and b. Assume that b>a. The cylinders are long enough so that we can neglect fringing of electric field at the ends.The outer one is earthed. Electric field at a point between the cylinders will be radial and its magnitude will depend on the distance from the central axis. Consider a Gaussian surface of length y and radius r such that a<r<b . Flux through the plane surface is zero because electric field and area vector are perpendicular to each other.

 

For the curved part,

$\begin{aligned} \phi &=\int \vec{E} \cdot d \vec{s}=\int E d s \\ &=E \int d s=E \cdot 2 \pi r y \end{aligned}$

Charge inside the gaussian surface, q=\frac{Qy}{L}

From Gauss's law   \phi=E 2 \pi r y=\frac{Q y}{L \varepsilon_{0}} \Rightarrow E=\frac{Q}{2 \pi \varepsilon_{0} L r}.

Potential difference: 

 \begin{aligned} V_{b}-V_{a}=-\int_{a}^{b} \bar{E} \cdot d \bar{r} &=-\int_{a}^{b} \frac{Q}{2 \pi \varepsilon_{0} L r} d r=-\frac{Q}{2 \pi \varepsilon_{0} L_{a}} \int_{a}^{b} \frac{1}{r} d r \\ \\ V_{a} &=\frac{Q}{2 \pi \varepsilon_{0} L} \ln \frac{b}{a} \quad\left(\text { since } V_{b}=0\right) \end{aligned}

Therefore, Capacitance of cylindrical capacitor is, 

 C=\frac{Q}{V_{\alpha}-V_{a}}=\frac{Q}{V_{a}}=\frac{2 \pi \varepsilon_{0} L}{\ln \left(\frac{b}{a}\right)}

 

 

Posted by

lovekush

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question