Spherical And Cylindrical Capacitors - Practice Questions & MCQ

Spherical capacitors: 

Spherical capacitors  has  two concentric spherical conducing shells of radii a and b, say b>a. The shell on the outer side is  earthed. We place a charge +Q on the inner shell. It will reside on the outer surface of the shell. A charge -Q will be induced on inner surface of outer shell. A charge +Q will flow from outer shell to the earth.

Consider a Gaussian spherical surface of radius $r$ such that $a<r<b$.
From Gauss's law, electric field at distance $r>a$ is

$
E=\frac{Q}{4 \pi \varepsilon_0 r^2}
$

The potential difference is :

$
V_b-V_a=-\int_0^b \bar{E} \cdot d \bar{r}=-\int_a^b \frac{Q}{4 \pi \varepsilon_0 r^2} d r
$

Since $V_b=0$, we have

$
V_b=\frac{Q}{4 \pi \varepsilon_0}\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{Q(b-a)}{4 \pi \varepsilon_0 a b}
$

Therefore, capacitance, $C=\frac{Q}{V_b-V_a}=\frac{Q}{V_a}=\frac{4 \pi \varepsilon a a b}{b-a}$

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, $q=\frac{Q y}{L}$
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)}
$