Module 2 : Electrostatics
Lecture 10 : Capacitance
The electric field at a distance $r$from the centre is calculated by using the Gaussian surface shown. The fileld is radial and is given by
 
\begin{displaymath}\vec E = \frac{Q}{4\pi\epsilon_0 r^2}\hat r\end{displaymath}
  The voltage drop between the shells is obtained by integrating the electric field along a radial path (the electric field being conservative, the path of integration is chosen as per our convenience) from the negative plate to the positrive plate.
 
\begin{eqnarray*} \phi &=& -\int_{R_2}^{R_1} \vec E\cdot\vec{dr} = \frac{Q}{4\p... ...\frac{Q}{4\pi\epsilon_0}\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \end{eqnarray*}
  The capacitance is
 
\begin{displaymath}C = \frac{Q}{\phi} = \frac{4\pi\epsilon_0 R_1 R_2}{R_2-R_1}\end{displaymath}
  Cylindrical Conductor :
  A cylindrical capacitor consists of two long coaxial conducting cylinders of length $L$and radii $R_1$and $R_2$. The electric field in the space between the cylinders may be calculated by Gauss Law, using a pillbox in the shape of a short coaxial cylinder of length $l\ll L$and radius $r$. Neglecting edge effects, the field is in the radial direction and depends only on the distance $r$from the axis.
   
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