• 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.

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

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

• 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.

• 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

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

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 \(C\mathcal E\).

• 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.

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