Module 2 : Electrostatics
Lecture 10 : Capacitance
where $q$is the charged enclosed by the pillbox, which is given in terms of the surface charge density $\sigma$on the inner cylinder by
 
\begin{displaymath}q = 2\pi R_1l\sigma \end{displaymath}
  The field at a distance $r$is given by
 
\begin{displaymath}\vec E = \frac{R_1\sigma}{\epsilon_0 r}\hat r \end{displaymath}
  The potential difference between the cylindrical conductors is
 
\begin{eqnarray*} \phi &=& -\int_{R_2}^{R_1} \vec E\cdot\vec{dr} = \frac{R_1\si... ...ac{dr}{r}\\ &=& \frac{R_1\sigma}{\epsilon_0}\ln\frac{R_2}{R_1} \end{eqnarray*}
  Substituting
 
\begin{displaymath}\sigma = \frac{Q}{2\pi R_1L}\end{displaymath}
   
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