Question

4. (15%) Find the electrostatic energy per unit length stored in a long cylindrical capacitor (inner radius a, outer radius b) with a potential difference between the plates equal to V, in two ways: (a) Find C and use W = CV? (b) Use W =) SEP dºx: you must find and integrate E. In this part, finish by expressing the charge per unit length 2 in terms of V to show that you get the same answer in (b) that you got in (a).

Answer 1

According to the Gauss's Law for Electrostatics,

Qenc Ē.ds Co

In this case of a long cylindrical capacitor, let its length be L, line charge density be $\lambda$ =(Q/L) and the medium between the plates is air with $\epsilon_r=1$ , i.e. $\mathbf{\epsilon =\epsilon _o}$

So, the Electric field intensity is given by

${E}*(2\pi rL)=\frac{Q}{\epsilon _{o}}\Rightarrow \mathbf{E=\frac{Q}{2\pi rL \epsilon_o}=\frac{\lambda }{2\pi r\epsilon _o}}$

The potential difference between the plates,

$\mathbf{V}=-\int_{b}^{a}\vec{E}.\vec{dl}=\frac{-Q}{2 \pi L \epsilon_o}\int_{b}^{a}\frac{1}{r}dr=\mathbf{\frac{Q}{2 \pi L \epsilon_o}\ln \frac{b}{a}=\frac{\lambda }{2\pi \epsilon _o}\ln\frac{b}{a}}$

The capacitance of the cylindrical capacitor (for the entire length) is

$\mathbf{C}=\frac{Q}{V}=\frac{Q}{\frac{Q}{2 \pi L \epsilon_o}\ln \frac{b}{a}}=\mathbf{\frac{2 \pi \epsilon_oL}{\ln\frac{b}{a}}}$

a)- The electrostatic energy per unit length stored in the capacitor,

$\mathbf{W}=\frac{1}{2L}CV^2=\frac{1}{2L}\frac{2\pi \epsilon _oL}{\ln\frac{b}{a}}(V)^{2}=\mathbf{\frac{\pi \epsilon _oV^2 }{\ln\frac{b}{a}}}$

b)- Now, as a matter of fact, the energy density associated with a capacitor is given by, $\mathbf{w}=\frac{Energy}{Volume}=\frac{\epsilon _o\left |\vec{E} \right |^2}{2}=\frac{\epsilon _o}{2}(\frac{Q}{2\pi r\epsilon _o L})^2=\mathbf{\frac{\lambda ^2}{8\pi^2 \epsilon_or^2 }}$

The volume of a cylinder is given by $Volume(V_o)=\pi r^2L\Rightarrow dV_o=\pi L*d(r^2)=(2\pi rL)dr$

Thus, the electrostatic energy per unit length stored in the capacitor,

$\mathbf{W}=\frac{1}{L}\int w.dV_o=\frac{1}{L}\int_{a}^{b}\frac{\lambda ^2}{8\pi^2 \epsilon_or^2 }\left (2\pi Lr \right )dr=\frac{\lambda ^2}{4\pi \epsilon _o}\int_{a}^{b}\frac{1}{r}dr=\mathbf{\frac{\lambda ^2\ln\frac{b}{a}}{4\pi \epsilon _o}}$

Representing $\lambda$ in terms of V as asked, we find that

$\Rightarrow \mathbf{W}=\frac{\lambda ^2\ln\frac{b}{a}}{4\pi \epsilon _o}=\frac{\ln\frac{b}{a}}{4\pi \epsilon _o}(\frac{2\pi \epsilon _oV}{\ln\frac{b}{a}})^2=\mathbf{\frac{\pi \epsilon _oV^2}{\ln\frac{b}{a}}}$

So, it can be easily observed that both the expressions for energy are identical.