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Work has to be done in charging a conductor against the force of repulsion by the already existing charges on it. The work is stored as a potential energy in the electric field of the conductor. Suppose a conductor of capacity C is charged to a potential V_{0} and let q_{0} be the charge on the conductor at this instant. The potential of the conductor when (during charging) the charge on it was q (< q_{0}) is,
Now, work done in bringing a small charge dq at this potential is,
Therefore, total work done in charging it from 0 to q_{0} is,
This work is stored as the potential energy,
Therefore,
Further by using q_{0} = CV_{0} we can write this expression also as,
In general if a conductor of capacity C is charged to a potential V by giving it a charge q, then
3.1 Energy Density of a Charged Capacitor
This energy is localized on the charges or the plates but is distributed in the field. Since in case of a parallel plate capacitor, the electric field is only between the plates, i.e., in a volume (A × d), the energy density
or
3.2 Calculation of Capacitance
The method for the calculation of capacitance involves integration of the electric field between two conductors or the plates which are just equipotential surfaces to obtain the potential difference V_{ab}. Thus,
Therefore,
3.3 Heat Generated :
(1) Work done by battery
W = QV
Q = charge flow in the battery
V = EMF of battery
(2) W = Ve (When Battery discharging)
W = Ve (When Battery charging)
(3) Q = CV (C = equivalent capacitance)
so W = CV × V = CV^{2}
Now energy on the capacitor _{}
Therefore, Energy dissipated in form of heat (due to resistance)
H = Work done by battery  {final energy of capacitor  initial energy of capacitor}
Ex.3 At any time S_{1} switch is opened and S_{2} is closed then find out heat generated in circuit.
Sol.
Charge flow through battery = Q_{f}  Q_{i}
= 2CV  CV = CV
H = (CV × 2V) 
Ex.4 (a) Find the final charge on each capacitor if they are connected as shown in the figure.
Sol. Initially
Finally let q charge flows clockwise then
Now applying KVL
5q + 50 + 5q  200 + 2q = 0
12 q  150 = 0 ⇒
so finally
(b) Find heat loss in the above circuit.
ΔH = Energy [initially  finally] on capacitor
When two capacitors C_{1} and C_{2} are connected as shown in figure
Before connecting the capacitors  
Parameter  I^{st} Capacitor  II^{nd} Capacitor 
Capacitance  C_{1}  C_{2} 
Charge  Q_{1}  Q_{2} 
Potential  V_{1}  V_{2} 
After connecting the capacitors  
Parameter  I^{st} Capacitor  II^{nd} Capacitor 
Capacitance  C_{1}  C_{2} 
Charge  
Potential  V_{1}  V_{2} 
(a) Common potential :
By charge conservation on plates A and C before and after connection.
Q_{1} + Q_{2} = C_{1}V + C_{2}V
⇒ =
(b)
(c) Heat loss during redistribution :
The loss of energy is in the form of Joule heating in the wire.
Derivation of above formulae :
Let potential of B and D is zero and common potential on capacitors is V, then at A and C it will be V.
C_{1}V + C_{2}V = C_{1}V_{1} + C_{2}V_{2}
=
H =
when oppositely charged terminals are connected then
Therefore, C_{1}V C_{2}V = C_{1}V_{1}  C_{2}V_{2}
Ex.5 Find out the following if A is connected with C and B is connected with D.
(i) How much charge flows in the circuit.
(ii) How much heat is produced in the circuit.
Sol. Let potential of B and D is zero and common potential on capacitors is V, then at A and C it will be V.
By charge conservation,
3V + 2V = 40 + 30
5V = 70 ⇒ V = 14 volt
Charge flow = 40  28 = 12 μC
Now final charges on each plate is shown in the figure.
(ii) Heat produced = × 2 × (20)^{2} × 3 × (10)^{2}  × 5 × (14)^{2}
= 400 150  490
= 550 490 = 60 mJ
Ex.6 Repeat above question if A is connected with D and B is connected with C.
Sol. Let potential of B and C is zero and common potential
on capacitors is V, then at A and D it will be V
2V + 3V = 10 ⇒ V = 2 volt
Now charge on each plate is shown in the figure.
Heat produced = 400 + 150  1
Therefore 2 × 5 × 4
= 550  10 = 540 μJ
Ex.7 Three capacitors as shown of capacitance 1mF, 2mF and 2mF are charged upto potential difference 30 V, 10 V and 15V respectively. If terminal A is connected with D, C is connected with E and F is connected with B. Then find out charge flow in the circuit and find the final charges on capacitors.
Sol. Let charge flow is q.
Now applying Kirchhoff's voltage low
 2q =  25
q = 12.5 mC
Final charges on plates
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