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Basic Concept

Learn basic concept and definition of capacitance, unit of capacitance. Practice formula to find capacitance of a parallel plate capacitor and electric field parallel plate capacitor, equivalent capacitance.

Definition of Capacitance

  • Consider conducting plates of a capacitor carrying charge of equal magnitude and opposite nature and potential difference between them is \(\Delta V\).

  • Experimentally, the relation between \(Q\) and \(\Delta V\) is given as,

\(Q\propto\Delta V\)

\(Q=C\Delta V\)

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

  • C is a constant known as capacitance of a capacitor and does not depend on Q or V.

Unit of Capacitance

  • Unit of capacitance is coulomb per volt or Farad.

\(1\,F=1\dfrac{C}{V}\)

  • Farad is very large unit of capacitance. Generally devices have capacitances in microfarad and picofarad.

\(1\mu F=10^{-6}F\)

\(1p F=10^{-12}F\)

Illustration Questions

Consider a capacitor having charge Q on one of its conducting plate. What will be the change in capacitance is charge is doubled?

A Doubled

B Half

C Remains the same

D Becomes one-fourth 

×

Capacitance (C) is a constant and does not depend on charge (Q) or voltage (V).

Consider a capacitor having charge Q on one of its conducting plate. What will be the change in capacitance is charge is doubled?

A

Doubled

.

B

Half

C

Remains the same

D

Becomes one-fourth 

Option C is Correct

Capacitance of Parallel Plates Capacitor

  • Consider two parallel metallic plates of equal area carrying a charge of equal magnitude and opposite nature are separated by a distance \(d\).
  • The surface charge density on each plate is \(\sigma\)

where \(\sigma=\dfrac{Q}{A}\)

  • When plates are placed very closed to each other (compared to length and width) electric field is assumed to be uniform.

\(E=\dfrac{\sigma}{\epsilon_0} =\dfrac{Q}{\epsilon_0A}\)   \(\left[\sigma=\dfrac{Q}{A}\right]\)

\(\Delta V=E.d\)         from Gauss Law

\(\Delta V=\dfrac{Qd}{\epsilon_0A}\)

  • Capacitance is given as \(C=\dfrac{Q}{\Delta V}\)

or, \(C=\dfrac{Q(\epsilon_0A)}{Qd}\)

or,  \(C=\dfrac{\epsilon_0A}{d}\)

  • Capacitance depends on shape and size of plates but is independent of Q and V.

Conclusion

The capacitance of a parallel plate capacitor is proportional to the area of the plates and inversely proportional to the separation between the plates.

 

 

Illustration Questions

A parallel plate capacitor having area (A) of its plates are separated by a distance (d). What will be the capacitance (C) if area is doubled?

A Remains same

B Doubled

C Half

D Becomes one-fourth

×

The capacitance of capacitor is given as

\(C=\dfrac{\epsilon_0A}{d}\)

When, 

\(A'=2A\)

New capacitance will be

\(C'=\dfrac{\epsilon_0(2A)}{d}=\dfrac{2\,\epsilon_0A}{d}\)

\(C'=2\,C\)

A parallel plate capacitor having area (A) of its plates are separated by a distance (d). What will be the capacitance (C) if area is doubled?

A

Remains same

.

B

Doubled

C

Half

D

Becomes one-fourth

Option B is Correct

Charge across Capacitor Plates

  • Consider a parallel plate capacitor connected across battery terminals.
  • The final charge Q is independent of initial charge on the plates.
  • Charge moves on the plates untill potential difference between plates becomes equal to battery's  \(e.m.f.\)

  • After some time,

Potential difference across the plates = Potential difference across the terminals of battery.

  • Charge on plate M be +Q and charge stored on plate N be –Q.
  • The magnitude of charge on the capacitor will be Q.

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

\(Q= C\Delta V\)

Note- Whatever be the initial charge, final charge remains the same.

Illustration Questions

Calculate the magnitude of charge on each plate of a parallel plate capacitor of capacitance \(C=7\,\mu F\) when connected across battery with emf  \( V=7\,V.\)

A \(60\,\mu C\)

B \(29\,\mu C\)

C \(49\,\mu C\)

D \(59\,\mu C\)

×

Charge on capacitor is given as

\(Q=CV\)

Given  \(C=7\,\mu F,\,V=7\,V\)

\(Q = 7\times10^{-6}\times7\)

\(Q = 49\times10^{-6}\)

\(Q = 49\,\mu C\)

Calculate the magnitude of charge on each plate of a parallel plate capacitor of capacitance \(C=7\,\mu F\) when connected across battery with emf  \( V=7\,V.\)

A

\(60\,\mu C\)

.

B

\(29\,\mu C\)

C

\(49\,\mu C\)

D

\(59\,\mu C\)

Option C is Correct

Electric field due to Parallel Plate Capacitor

  • Consider two parallel metallic plates of equal area carrying a charge of equal magnitude and opposite nature are separated by a distance \(d\).
  • The surface charge density on each plate is \(\sigma\)

where \(\sigma=\dfrac{Q}{A}\)

  • When plates are placed very closed to each other (compared to length and width) electric field is assumed to be uniform.

 

  • Since distance between plates is small, it can be considered as single conductor.

By Gauss's law,

Flux through Gaussian surface

\(\phi _E=\oint\,E.dA=\dfrac{q_{in}}{\epsilon_0}\)

or,  \(\phi _E=2 EA\)   ....(1)

Also,   \(\phi _E=\dfrac{q_{in}}{\epsilon_0}\) ....(2)

From (1) and (2)

\(2EA=\dfrac{\sigma A}{\epsilon_0}\)   \([q_{in}=\sigma\times A]\)

or, \(E_+=\dfrac{\sigma}{2\epsilon_0}\)

\(E_+=\dfrac{Q}{2\epsilon_0A}\)

Similarly for negative charge plate

 \(E_-=\dfrac{Q}{2\epsilon_0A}\)

Total field =\(E_++E_-\)

Total field \(=\dfrac{Q}{2\epsilon_0A}+\dfrac{Q}{2\epsilon_0A}=\dfrac{Q}{\epsilon_0A}\)

 

Illustration Questions

Calculate electric field between plates of parallel plate capacitor having area  \(A=1\,m^2\) and charge on each plate \(Q=8.85\,nC.\) 

A \(10^2\,N/C\)

B \(10^5\,N/C\)

C \(10^4\,N/C\)

D \(10^3\,N/C\)

×

Electric field due to parallel plate capacitor is given as 

\(E=\dfrac{Q}{\epsilon_0A}\)

Given : \(Q=8.85\,nC\,\,,\,\,A=1\,m^2\)

\(E=\dfrac{8.85\times 10^{-9}}{8.85\times 10^{-12}\times1}\)

\(E=10^3\,N/C\)

Calculate electric field between plates of parallel plate capacitor having area  \(A=1\,m^2\) and charge on each plate \(Q=8.85\,nC.\) 

A

\(10^2\,N/C\)

.

B

\(10^5\,N/C\)

C

\(10^4\,N/C\)

D

\(10^3\,N/C\)

Option D is Correct

Equivalent Capacitance due to Parallel Combination of Capacitors

  • Two capacitors connected in parallel are joined from top to top and from bottom to bottom.
  • Top electrode of both the capacitors are connected by a conducting wire so that they are at same potential.
  • Similarly, bottom electrode of both the capacitors are connected by a conducting wire so that they are at same potential.
  • Thus, it means two capacitors connected in parallel, have same potential difference.
  • Consider a circuit in which two capacitors (C1 and C2) are connected in parallel with a battery.
  • The potential difference across each capacitor is assumed to be \(\Delta V_c\).
  • Charge stored by capacitor C1

?\(Q_1=C_1\,\Delta V_c\)

  • Charge stored by capacitor C2

\(Q_2=C_2\,\Delta V_c\)

 

 

  • Now, the above circuit is replaced with new circuit having single capacitor in place of two capacitors having total charge \(Q=Q_1+Q_2\)

  • The voltage across capacitor is \(\Delta V_c\) . 
  • Thus, the capacitance of capacitor is given as 

\(C_{eq}=\dfrac{Q}{\Delta V_c}\)

or,  \(C_{eq}=\dfrac{Q_1+Q_2}{\Delta V_c}\)

or,  \(C_{eq}=\dfrac{Q_1}{\Delta V_c}+\dfrac{Q_2}{\Delta V_c}\)

or,  \(C_{eq}=C_1+C_2\)

Conclusion :

If two or more capacitors are connected in parallel then their equivalent capacitance is given as

\(C_{eq}=C_1+C_2+C_3+...\)

 

 

Illustration Questions

Determine equivalent capacitance for the given circuit.

A \(3\,\mu F\)

B \(5\,\mu F\)

C \(7\,\mu F\)

D \(8\,\mu F\)

×

Equivalent capacitance for parallel combination is sum of capacitance of all capacitors.

image

Equivalent capacitance 

\(C_{eq}=3\,\mu F+1\,\mu F+2\,\mu F+2\,\mu F\)

\(C_{eq}=8\,\mu F\)

image

Determine equivalent capacitance for the given circuit.

image
A

\(3\,\mu F\)

.

B

\(5\,\mu F\)

C

\(7\,\mu F\)

D

\(8\,\mu F\)

Option D is Correct

Calculation of Charge Provided by Battery

Case 1 :

  • Consider a capacitor, initially having some charge on its plate of magnitude q.

 

 

  • A battery with emf \(\mathcal E\) is connected with the capacitor by throwing the switch from A to B.
  • As battery is connected, battery charges the capacitor upto the maximum value i.e., \(C\mathcal E\) .

\(Q_{max}=C\mathcal E\)  [\(\therefore\)Final charge on the capacitor after connection of battery]

  • Charge provided by the battery

\(\Delta Q =Q_{max}-q\)      [\(\therefore\) \(\Delta Q=Q_f-Q_i\)

\(\Delta Q=C\mathcal E-q\)

Case 2 :

Initially case configuration is as shown in figure.

  • On moving switch from A to B, for positive plate charge provided by battery to charge upto maximum value i.e., \(\,C\mathcal E\)

\(\Delta Q = Q_{max} -(-q)\)

\(\Delta Q=C\mathcal E+q\)

Note- Charge provided by positive terminal of battery is  \(C\mathcal E+q\)  as battery first levels \(-q\) to zero and then charges the plate upto \(C\mathcal E\).

Illustration Questions

A parallel plate capacitor with capacitance \(C=7\,\mu F\) having charge \(Q=3\,\mu C\) on plate A is connected with a battery of  \(\mathcal E=1\,V.\) Calculate the charge provided by the battery.

A \(4\,\mu C\)

B \(3\,\mu C\)

C \(8\,\mu C\)

D \(10\,\mu C\)

×

On moving switch from A to B, battery starts charging capacitor upto  \(Q_{max}\).

image

Maximum charge on capacitor

\(Q_{max}=C\mathcal E\)

Given : \(C=7\,\mu F,\,\,\,\mathcal E=1\,V\)

\(Q_{max}=7\,\mu F\times 1\,V\)

\(Q_{max}=7\,\mu C\)

image

Charge provided by battery

\(\Delta Q=Q_{max}-Q\)

\(\Delta Q=7\,\mu C-3\,\mu C\)

\(\Delta Q=4\,\mu C\)

image

A parallel plate capacitor with capacitance \(C=7\,\mu F\) having charge \(Q=3\,\mu C\) on plate A is connected with a battery of  \(\mathcal E=1\,V.\) Calculate the charge provided by the battery.

image
A

\(4\,\mu C\)

.

B

\(3\,\mu C\)

C

\(8\,\mu C\)

D

\(10\,\mu C\)

Option A is Correct

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