Physics > Capacitors > 6.0 Capacitance of a cylindrical capacitor
Capacitors
1.0 Introduction
2.0 Different types of capacitors and its capacitance.
3.0 Parallel Plate Capacitor
3.1 When equal and opposite charges placed on plates
3.2 When unequal charges are placed on the two plates
4.0 Capacitance of spherical conductor
5.0 Capacitance of a earthed sphere by a concentric spherical shell
6.0 Capacitance of a cylindrical capacitor
7.0 Mechanical force on the charged conductor
8.0 Redistribution of Charge
9.0 Dielectrics (Insulators) and Polarization
9.1 Effect of Dielectric
9.2 Capacitance of a Capacitor Partially Filled with Dielectric
9.3 Quantities after inserting dielectric in a capacitor (fully)
10.0 Combination of capacitors
11.0 Energy Density ($u$)
12.0 $R$-$C$ Circuits
13.0 Method of Finding Equivalent Capacitance
14.0 Some important concepts
15.0 Van De Graaff Generator
6.1 Working method to find capacitance of any capacitor
3.2 When unequal charges are placed on the two plates
9.2 Capacitance of a Capacitor Partially Filled with Dielectric
9.3 Quantities after inserting dielectric in a capacitor (fully)
- To find capacitance of a conductor first we will find voltage (potential difference ($V$)) of the system in form of charge $q$.
- It may find by integration of electrical field $$V = - \int\limits_a^b {\overrightarrow E .\overrightarrow {dr} } $$ or may by standard formulas.
- After finding $V$ in terms of $q$, compare with generator $q=CV$ and find the value of $C$.
- We can find capacitance for standard by using formulas also. i.e.,
1. For parallel plate capacitor: $$C = \frac{{A{\varepsilon _ \circ }}}{d}$$
2. For spherical shell capacitor: $$C = 4\pi {\varepsilon _ \circ }\left( {\frac{{ab}}{{b - a}}} \right)$$
3. For cylindrical capacitor: $C$ per unit length is given by $$\frac{{2\pi {\varepsilon _ \circ }}}{{\ln (b/a)}}$$
Example: Three parallel metallic plates each of area $A$ are kept as shown in figure and charges $q_1$, $q_2$ and $q_3$ are given to them. Find the resulting charge distribution on the six surfaces, neglecting edge effects as usual.
Solution: In the starting let first calculate total charge on the system that is $q_1+q_2+q_3$
Now, divide it by $2$ and put on two extreme surfaces ($a$ and $f$) so $${q_a} = \frac{{{q_1} + {q_2} + {q_3}}}{2} = {q_f}$$
Now total charge on plate $P$ should be $q_1$ so$${q_b} = {q_1} - \left( {\frac{{{q_1} + {q_2} + {q_3}}}{2}} \right) = \frac{{{q_1} - {q_2} - {q_3}}}{2} = - {q_c}$$ Because surface $b$ and $c$ is forming a capacitor. Now total charge on plate $Q$ should be $q_2$ so,$$q_d=q_2-q_c$$$${q_d} = {q_2} + \frac{{{q_1} - {q_2} - {q_3}}}{2} = \frac{{{q_1} + {q_2} - {q_3}}}{2} = - {q_e}$$
Example: Three concentric shells $A$, $B$ and $C$ of radii $a$, $b$ and $c$ are as shown in figure. Find the capacitance of the assembly between $A$ and $C$.
Solution: Consider a positive charge $q$ on $A$ and a negative charge $-q$ on $C$. Find the potential difference 
$V$ between $A$ and $C$.The desired capacitance will be $$C = \frac{q}{V}$$ Using the generator principle,$$V = {V_A} - {V_C} = \frac{q}{{4\pi {\varepsilon _ \circ }}}\left( {\frac{1}{a} - \frac{1}{c}} \right) = \frac{q}{{4\pi {\varepsilon _ \circ }}}\left( {\frac{{c - a}}{{ac}}} \right)$$ The desired capacitance is,$$C = \frac{q}{V} = 4\pi {\varepsilon _ \circ }\left( {\frac{{ac}}{{a - c}}} \right)$$
$V$ between $A$ and $C$.The desired capacitance will be $$C = \frac{q}{V}$$ Using the generator principle,$$V = {V_A} - {V_C} = \frac{q}{{4\pi {\varepsilon _ \circ }}}\left( {\frac{1}{a} - \frac{1}{c}} \right) = \frac{q}{{4\pi {\varepsilon _ \circ }}}\left( {\frac{{c - a}}{{ac}}} \right)$$ The desired capacitance is,$$C = \frac{q}{V} = 4\pi {\varepsilon _ \circ }\left( {\frac{{ac}}{{a - c}}} \right)$$