Learn steps to capacitance of a capacitor calculation, practice to calculation of potential difference between charged shells and spherical & cylindrical capacitor.

- Consider two oppositely charged conductors having magnitude of charge Q with simple geometry.
- To calculate capacitance following steps are followed :

- Identify two conductors, if there is any one conductor then it also has capacitance because we can imagine the second conductor at infinity, having the same charge but opposite nature.
- Calculate electric field E by using Gauss's Law.
- Calculate potential difference using electric field \(|\Delta V|=\int\;E\cdot ds\)
- Capacitance of a capacitor can be calculated as \(C=\dfrac {Q}{|\Delta V|}\)

A Charge (Q)\(\rightarrow\) Electric field (E)\(\rightarrow\) Potential difference(\(\Delta V\)) \(\rightarrow\)Capacitance, \(C=\dfrac {Q}{\Delta V}\)

B Charge (Q)\(\rightarrow\) Potential difference(\(\Delta V\)) \(\rightarrow\) Electric field (E)\(\rightarrow\)Capacitance, \(C=\dfrac {\Delta V}{E}\)

C Charge (Q)\(\rightarrow\) Potential difference(\(\Delta V\)) \(\rightarrow\) Electric field (E)\(\rightarrow\)Capacitance, \(C=\dfrac {Q}{E}\)

D None of these

- Consider a spherically charged conductor having radius R and charge +Q.
- The electric potential of a spherically charged conductor is given as \(V=\dfrac {KQ}{R}\)

**Step 1 : **Consider a spherical conducting shell of radius b, having charge –Q concentric with a smaller conducting sphere of radius a, having charge +Q.

**Step 2: **Electric field outside the sphere is in radial direction (if radius is r)

\(E=\dfrac {KQ}{r^2}\)

**Step 3 : **Potential difference between a conducting sphere and a conducting shell, is given as

\(V_b-V_a=-\int\limits_a^b\vec E\cdot\vec dr\)

or, \(V_b-V_a=-\int\limits_a^b\dfrac {KQ}{r^2}\;dr\)

or, \(V_b-V_a=-KQ\int\limits_a^b\dfrac {1}{r^2}\;dr\)

or, \(V_b-V_a=-KQ \left [ -\dfrac {1}{r} \right]_a^b\)

or, \(V_b-V_a=KQ \left [ \dfrac {1}{b}-\dfrac {1}{a} \right]\)

The magnitude of the potential difference is

\(\Delta V \,=\,\, |V_b-V_a|=\dfrac {KQ(b-a)}{ab} \)

**Step 4 : **Capacitance of the body is given as

\(C=\dfrac {Q}{\Delta V}\)

or, \(C=\dfrac {Q}{KQ\left ( \dfrac {b-a}{ab}\right)}\)

or, \(C=\dfrac {ab}{K(b-a)}\)

A Charge (Q)\(\rightarrow\) Potential difference(\(\Delta V\)) \(\rightarrow\)Capacitance, \(C=\dfrac {\Delta V}{Q}\)

B Charge (Q)\(\rightarrow\)Electric field (E)\(\rightarrow\) Potential difference(\(\Delta V\)) \(\rightarrow\)Capacitance, \(C=\dfrac {Q}{\Delta V}\)

C Charge (Q)\(\rightarrow\)Potential difference(\(\Delta V\)) \(\rightarrow\)Electric field (E)\(\rightarrow\) Capacitance, \(C=\dfrac {Q}{\Delta V}\)

D None of these

- Consider two oppositely charged conductors having magnitude of charge Q with simple geometry.
- To calculate capacitance following steps are followed:

- Identify two conductors, if there is any one conductor then it also has capacitance because we can imagine the second conductor at infinity, having the same charge but opposite in nature.
- Calculate electric field E by using Gauss's Law.
- Calculate potential difference using electric field \(|\Delta V|=\int\;E\cdot ds\)

- Consider a spherically charged conductor having radius R and charge +Q.

- The electric potential of a spherically charged conductor is given as \(V=\dfrac {KQ}{R}\)

**Step 1 : **Consider a spherical conducting shell of radius b, having charge –Q concentric with a smaller conducting sphere of radius a, having charge +Q.

**Step 2: **Electric field outside the sphere is in radial direction (if radius is r)

\(E=\dfrac {KQ}{r^2}\)

**Step 3 : **Potential difference between a conducting sphere and a conducting shell, is given as

\(V_b-V_a=-\int\limits_a^b\vec E\cdot\vec dr\)

or, \(V_b-V_a=-\int\limits_a^b\dfrac {KQ}{r^2}\;dr\)

or, \(V_b-V_a=-KQ\int\limits_a^b\dfrac {1}{r^2}\;dr\)

or, \(V_b-V_a=-KQ \left [- \dfrac {1}{r} \right]_a^b\)

or, \(V_b-V_a=KQ \left [ \dfrac {1}{b}-\dfrac {1}{a} \right]\)

The magnitude of the potential difference is

or, \(\Delta V=\,\,|V_b-V_a|=\dfrac {KQ(b-a)}{ab} \)

**Step 4 : **Capacitance of the body is given as

\(C=\dfrac {Q}{\Delta V}\)

or, \(C=\dfrac {Q}{KQ\left ( \dfrac {b-a}{ab}\right)}\)

or, \(C=\dfrac {ab}{K(b-a)}\)

The electric field lines for the inner conductor will be radially outward.

- Assume that the outer sphere is made of infinitely large radius,

means \(b\rightarrow\infty\)

- Also, the electric field lines for outer spherical conductor will be radially outward.
- If the outer radius (b) of sphere approaches \(\infty\), then capacitor becomes isolated spherical conductor.
- The capacitance of isolated spherical conductor is given as

\(C= \displaystyle\lim_{b\to\infty} \dfrac {ab}{K(b-a)}\)

or, \(C=\displaystyle\lim_{b\to\infty}\dfrac {a}{K\left ( 1-\dfrac {a}{b}\right)}\) \(\left [ \displaystyle\lim_{b\to\infty}\dfrac {a}{b}=0 \right]\)

\(C=\dfrac {a}{K}\)

\(C=4\pi\epsilon_0a\)

This is the capacitance for isolated sphere.

A \(20\pi\epsilon_0\,F\)

B \(5\pi\epsilon_0\,F\)

C \(10\pi\epsilon_0\,F\)

D \(\pi\epsilon_0\,F\)

- Consider a cylindrical conductor of radius 'a', having charge Q, is co-axial with a cylindrical shell of radius 'b'.
- The length of cylindrical capacitor is \(\ell\).

**Step 1: **The electric field at a distance 'r' from a line charge of infinite length is given as \(E=\dfrac {\lambda }{2\pi\epsilon_0r}\)

where \(\lambda\) = charge per unit length

**Step 2 : ** Potential difference in terms of electric field is given as

\(\Delta V = -\int\limits_a^b\;\vec E\cdot\vec ds\)

\(V_b-V_a= -\int\limits_a^b\;\vec E\cdot\vec ds\)

- Assume \(\ell>>>\) radius (a or b)
- Then the cylindrical capacitor will appear to be line charge of infinite length and electric field is given as \(E=\dfrac {\lambda}{2\pi\epsilon_0r}\)
- Potential difference between two conductors, is given as

\(V_b-V_a=-\int\limits_a^b \dfrac {\lambda}{2\pi\epsilon_0r}dr\)

\(\Rightarrow V_b-V_a=-\dfrac {\lambda}{2\pi\epsilon_0}\int\limits_a^b\dfrac {1}{r} dr\)

\(\Rightarrow V_b-V_a=-\dfrac {\lambda}{2\pi\epsilon_0} [\ell n\;r]_a^b\)

\(\Rightarrow V_b-V_a=-\dfrac {\lambda}{2\pi\epsilon_0} [\ell n\,b-\ell n\,a]\)

\(\Rightarrow V_b-V_a=-\dfrac {\lambda}{2\pi\epsilon_0} \left [\ell n(\dfrac {b}{a})\right]\)

\(\left [\lambda=\dfrac {Q}{\ell}\text{where }\lambda \,\,\text{is charge per unit length} \right]\)

\(\Rightarrow V_b-V_a=-\dfrac {Q}{2\pi\epsilon_0\ell}\ell n \left( \dfrac {b}{a} \right)\)

The magnitude of the potential difference is

\( \Delta V=|V_b-V_a|=\dfrac {Q}{2\pi\epsilon_0\ell}\ell n \left( \dfrac {b}{a} \right)\)

**Step 3 : **Capacitance of cylindrical capacitor is given as

\(C=\dfrac {Q}{\Delta V}=\dfrac {Q}{V_b-V_a}\)

\(\Rightarrow C=\dfrac {Q} {\dfrac {Q}{2\pi\epsilon_0 \ell}\ell n\left (\dfrac {b}{a}\right)}\)

\(\Rightarrow C=\dfrac {\ell}{2K\ell n\left (\dfrac {b}{a}\right)}\)

A \(\dfrac{1}{\ell n\,(4)} nF\)

B \(\dfrac{2}{\ell n\,(5)} nF\)

C \(\dfrac{1}{\ell n\,(2)} nF\)

D \(\dfrac{2}{\ell n\,(2)} nF\)

A \(\dfrac {2}{1}\)

B \(\dfrac {3}{1}\)

C \(\dfrac {4}{3}\)

D \(\dfrac {6}{5}\)