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Calculation Of Capacitance

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

Steps to Calculate Capacitance of Capacitor

  • Consider two oppositely charged conductors having magnitude of charge Q with simple geometry.
  • To calculate capacitance following steps are followed :
  1. 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.
  2. Calculate electric field E by using Gauss's Law.
  3. Calculate potential difference using electric field \(|\Delta V|=\int\;E\cdot ds\)
  4. Capacitance of a capacitor can be calculated as \(C=\dfrac {Q}{|\Delta V|}\)

Illustration Questions

Choose the correct sequence of steps to calculate capacitance for parallel plate capacitor having charge + Q and –Q respectively.

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

×

Determine the charge (Q) on parallel plate capacitor.

Calculate electric field (E) by using Gauss's Law.

Calculate potential difference by using electric field \(|\Delta V|=\int E \cdot ds\)

Then, calculate capacitance of capacitor by using \(C=\dfrac {Q}{|\Delta V|}\)

Hence, option (A) is correct.

Choose the correct sequence of steps to calculate capacitance for parallel plate capacitor having charge + Q and –Q respectively.

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

Option A is Correct

Calculation of Capacitance of a Spherically Charged Conductor

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

 

Stepwise Calculation of Capacitance for Spherical Capacitor

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, \(C=\dfrac {ab}{K(b-a)}\)

 

 

 

Illustration Questions

Choose the correct sequence of steps to calculate capacitance for a spherical capacitor.

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

×

Determine the charge (Q) on a spherical capacitor.

Calculate the electric field (E) by using Gauss's Law \(E=\dfrac {KQ}{r^2}\)

Calculate potential difference by using electric field \(\Delta V=-\int\vec E\cdot\vec dr\)

Then, calculate capacitance of spherical capacitor by using \(C=\dfrac {Q}{\Delta V}\)

Hence, option (B) is correct.

Choose the correct sequence of steps to calculate capacitance for a spherical capacitor.

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

Option B is Correct

Calculation of Potential Difference between Charged Shells

  • Consider two oppositely charged conductors having magnitude of charge Q with simple geometry.
  • To calculate capacitance following steps are followed:
  1. 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.
  2. Calculate electric field E by using Gauss's Law.
  3. Calculate potential difference using electric field \(|\Delta V|=\int\;E\cdot ds\)

Illustration Questions

A spherical capacitor of conducting shell of radius (b) = 5 m having charge (–Q) =\(-3\,nC\) is concentric with a smaller conducting sphere of radius (a) = 2m and charge (Q) = \(+3\,nC\). Calculate the potential difference between the shells.

A 9 V

B 10 V

C 8.1 V

D 16 V

×

Potential difference between a conducting sphere and a conducting shell is given as 

\(V_b-V_a=\dfrac {KQ(b-a)}{ab}\)

image

Given : a = 2m, b = 5m, \(Q= 3\,nC\)

\(\Rightarrow\) \(V_b-V_a=\dfrac {9\times10^9\times 3\times10^{-9}\times(5-2)}{5\times2}\)

\(\Rightarrow V_b-V_a=\dfrac {9\times10^9\times 9\times10^{-9}}{10}\)

\(\Rightarrow V_b-V_a=\) 8.1 V

image

Hence, option (C) is correct.

image

A spherical capacitor of conducting shell of radius (b) = 5 m having charge (–Q) =\(-3\,nC\) is concentric with a smaller conducting sphere of radius (a) = 2m and charge (Q) = \(+3\,nC\). Calculate the potential difference between the shells.

image
A

9 V

.

B

10 V

C

8.1 V

D

16 V

Option C is Correct

Calculation of Capacitance of an Isolated Spherical Conductor

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

 

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

Stepwise Calculation of Capacitance for Spherical Capacitor

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, \(C=\dfrac {ab}{K(b-a)}\)

 

For Isolated Spherical Conductor

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.

Illustration Questions

Calculate the value of capacitance of an isolated spherical conductor of radius a = 5 m.

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

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

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

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

×

Capacitance for an isolated sphere is given as \(C=4\pi\epsilon_0a\)

Given : a = 5m, 

\(C=4\pi\epsilon_0\times 5\)

\(C=20\pi\epsilon_0\,F\)

Hence , option (A) is correct.

Calculate the value of capacitance of an isolated spherical conductor of radius a = 5 m.

A

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

.

B

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

C

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

D

\(\pi\epsilon_0\,F\)

Option A is Correct

Calculation of Capacitance of Cylindrical Capacitor

  • 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

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

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

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

Illustration Questions

Calculate capacitance of a cylindrical capacitor of length \(\ell\)= 18 m, having inner radius and outer radius of a = 2 cm and b = 4 cm respectively.

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

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

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

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

×

The capacitance of a cylindrical capacitor is given as

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

Given : \(\ell\)=18 m, a = 2 cm, b = 4cm

\(C=\dfrac {18}{2\times9\times10^9 \ell n\,(4/2)}\)

\(\Rightarrow C=\dfrac {10^{-9}}{\ell n\,(2)}\)

\(\Rightarrow C=\dfrac {1}{\ell n(2)}\;n F\)

Calculate capacitance of a cylindrical capacitor of length \(\ell\)= 18 m, having inner radius and outer radius of a = 2 cm and b = 4 cm respectively.

A

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

.

B

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

C

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

D

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

Option C is Correct

Illustration Questions

Calculate the ratio of the inner and outer radius of the cylindrical capacitor of capacitance \(C=\dfrac {1}{\ell n\;(2)}nF\) and length  \(\ell\)= 18 m.

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

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

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

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

×

The capacitance of a cylindrical capacitor is given as \(C=\dfrac {\ell}{2K\;\ell n\left ( \dfrac {b}{a}\right)}\)

 

Given : \(C=\dfrac {1}{\ell n(2)}nF\), \(\ell\)= 18 m

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

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

\(\Rightarrow \left ( \dfrac {b}{a} \right) =e^{\ell/2KC}\)

\(\Rightarrow \left ( \dfrac {b}{a} \right) =e^\left ({\dfrac {18\;\ell n(2)}{2\times9\times10^9\times1\times10^{-9}}}\right)\)

\(\Rightarrow \left ( \dfrac {b}{a} \right) =\dfrac {2}{1}\)

Calculate the ratio of the inner and outer radius of the cylindrical capacitor of capacitance \(C=\dfrac {1}{\ell n\;(2)}nF\) and length  \(\ell\)= 18 m.

A

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

.

B

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

C

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

D

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

Option A is Correct

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