Informative line

### Energy Stored In A Capacitor

Learn how to find the total energy stored in each capacitor of a circuit and calculate the work done by battery when the initial charge present on capacitor and the change in potential energy when the polarity of battery is reversed.

# Energy Stored in Capacitor

• Consider a parallel plate capacitor of area A, having charge "+ Q" on one plate and charge "– Q" on another plate.
• Electric field due to positive plate of capacitor is given as,

$$E_+ = \dfrac{\sigma}{2\epsilon _o} = \dfrac{Q}{2A\epsilon_ o}$$

• Force on negative plate of capacitor due to field of positive plate is given as,

$$F=-QE$$

$$F= -Q.\dfrac{Q}{2A\epsilon _o}$$

$$F = \dfrac{–Q^{2}}{2A\epsilon_o}$$

• Magnitude of force is given as,

$$F = \dfrac{Q^{2}}{2A\epsilon_o}$$ ( Force of attraction )

• Force of attraction is the force by which both the plates attract each other.
• Initially these plates are kept quite close to each other, i.e., separation between the plates is very small.
• Now one plate is kept fixed and other plate is moved away slowly till the separation between the plates increases from zero to "d".
• Since, there is a force of attraction between plates of capacitor so, work is done to separate them upto distance "d".

$$W = Fd$$

$$W = \dfrac{Q^{2}d}{2A\epsilon_0}$$

or      $$W = \dfrac{Q^{2}}{2 C}$$

• This work  done to separate the plate is stored as the increase of energy in the system which is same as the amount of energy stored in capacitor.

$$U = \dfrac{Q^{2}}{2 C}$$

• Consider a capacitor of any type to be in charging process.
• At any instant, the instantaneous value of charge on the plates of capacitor are +q and –q respectively.  • The instantaneous value of potential difference is

$$\Delta V = \dfrac{q}{C}$$

• At the next instant of time, $$dq$$  amount of charge is transferred from negative plate to positive plate.
• Work must be done to transfer $$dq$$  amount of charge from negative plate to positive plate,i.e., to move charge from plate at lower potential to plate at higher potential.

$$dW=\dfrac{q} {C}dq$$

•  Since, the capacitor can be charged up to Q so, the total work done in charging is

$$\int\limits ^W_0dW = \int\limits^Q_0 \dfrac {q}{C} dq$$

or       $$W = \dfrac{Q^{2}}{2 C}$$

• This work done in charging the capacitor up to Q appears as energy stored in capacitor.

$$U = \dfrac{Q^{2}}{2 C}$$

#### Consider a parallel plate capacitor, having charge Q and capacitance C. What will happen if the separation between plates is doubled, given the potential difference across capacitor is $$\Delta$$V? ( Assume capacitor is not connected to battery )

A Charge is doubled

B Charge is halved

C Potential difference is doubled

D Potential difference is halved

×

The capacitance of a capacitor is given as

$$C = \dfrac {\epsilon_0 A}{d}$$

where ,C = Capacitance

A = Area of plates

d = Separation between plates

When separation between plates is doubled,

$$d^{'} = 2 d$$

then, the new capacitance is given as,

$$C^{'} = \dfrac {\epsilon_0 A}{2d}$$

or, $$C^{'} = \dfrac {C}{2}$$

Now charge on capacitor is given as,

$$Q=C^{'}V$$

Since, the charge remains constant so,

$$C\propto \dfrac {1}{V}$$

Hence, when capacitance is halved, the potential difference is doubled.

### Consider a parallel plate capacitor, having charge Q and capacitance C. What will happen if the separation between plates is doubled, given the potential difference across capacitor is $$\Delta$$V? ( Assume capacitor is not connected to battery )

A

Charge is doubled

.

B

Charge is halved

C

Potential difference is doubled

D

Potential difference is halved

Option C is Correct

# Energy Stored in each Capacitor of a Circuit

• Consider a circuit having two capacitors in series of capacitance C1 and C2 respectively, are connected with a battery of e.m.f. $$\mathcal{E}$$ as shown in figure.  • Since capacitors C1 and C2 are connected in series , so charge stored on both the capacitors, is same, i.e., Q.
• Energy stored in capacitor C1

$$U_1 =\dfrac {Q^{2}}{2C_1}$$

• Energy stored in capacitor C2

$$U_2 =\dfrac {Q^{2}}{2C_2}$$

• Then, total energy of the system is given as

U = U1 + U2

$$\Rightarrow U = \dfrac {Q^{2}}{2C_1} + \dfrac {Q^{2}}{2 C_2}$$

$$\Rightarrow U = Q^{2} ( \dfrac {1}{2C_1} + \dfrac{1}{2C_2})$$

$$\Rightarrow U = \dfrac {Q^{2}}{2} \left (\dfrac {1}{C_1} + \dfrac{1}{C_2} \right)$$

#### Two capacitors of capacitance C1 = 12 $$\mu$$F and C2 = 6 $$\mu$$F, are connected in series with $$\mathcal{E}$$= 3 V battery. Calculate the energy stored in each capacitor.

A  $$8\,\mu J,\,12\,\mu J$$

B $$6\,\mu J,\,12\,\mu J$$

C $$4\,\mu J,\,12\,\,\mu J$$

D $$12\,\mu J,\,3\,\mu J$$

×

Equivalent capacitance when C1 and C2 are connected in series

$$\dfrac {1}{C_{eq}} = \dfrac {1}{C_1} + \dfrac {1}{C_2}$$

or      $$\dfrac {1}{C_{eq}} = \dfrac {1}{12 × 10^{-6}} + \dfrac {1}{6 × 10^{-6}}$$

or    $$\dfrac {1}{C_{eq} } = \dfrac {1}{4 \,\mu F}$$

or, $$C_{eq} = 4\, \mu F$$ Charge on each capacitor connected in series, is same and given as

$$Q=C\mathcal{E}$$

or    $$Q = 4 × 10 ^{-6} × 3$$

or   $$Q = 12 \,$$$$\mu C$$ Energy stored in capacitor C1

or,   $$U_1 = \dfrac {Q^{2}}{2 C_1}$$

or,   $$U_1 = \dfrac {(12 × 10 ^{-6})^{2}}{2 × (12 × 10 ^{-6})}$$

or,  $$U_1 = 6 \,\mu J$$ Energy stored in capacitor C2

or,   $$U_2 = \dfrac {Q^{2}}{2 C_2}$$

or,   $$U_2 = \dfrac {(12 × 10 ^{-6})}{2 × 6 × 10 ^{-6}}^{2}$$

or,  $$U_2 = 12 \,\mu J$$ ### Two capacitors of capacitance C1 = 12 $$\mu$$F and C2 = 6 $$\mu$$F, are connected in series with $$\mathcal{E}$$= 3 V battery. Calculate the energy stored in each capacitor.

A

$$8\,\mu J,\,12\,\mu J$$

.

B

$$6\,\mu J,\,12\,\mu J$$

C

$$4\,\mu J,\,12\,\,\mu J$$

D

$$12\,\mu J,\,3\,\mu J$$

Option B is Correct

# Work Done by Battery when its Polarity is Interchanged

• Consider a capacitor having capacitance C is connected to a battery with e.m.f. $$\mathcal{E}$$ as shown in figure.  • Initially, the capacitor is uncharged,  i.e., Qi = 0.
• When capacitor is connected to battery, the capacitor gets fully charged up to  maximum value (Qmax).

Qf = Qmax = $$C\mathcal{E}$$

• Hence, the charge flow through battery is

$$\Delta Q = Q_f – Q_i$$

or,   $$\Delta Q = Q_{max} – 0$$

or,   $$\Delta Q = Q_{max} =C \mathcal{E}$$

• Work done by battery,

W1 = $$\Delta Q × \mathcal{E} = C\mathcal{E} ^{2}$$

• Now, if the polarity of battery is reversed.
• Before reversing the polarity of battery, charge on negative plate is – Q ,

Q'i = – Q  • After reversing the battery charge on negative plate is + Q.  • Final charge on capacitor,

Q'= + Q = + $$C \mathcal{E}$$

• Hence, charge flow through battery is

$$\Delta$$Q' = Q'f – Q'i

or,     $$\Delta$$Q' = + Q – (– Q)

or,     $$\Delta$$Q' = 2 Q

• Hence, Work done by battery

W2 = $$\Delta$$Q' × $$\mathcal{E}$$ $$= 2 \,Q ×\mathcal{E}$$

= + $$2 C\mathcal{E}^{2}$$

#### An uncharged capacitor having capacitance $$C = 3 \,\mu F$$, is connected to a battery $$\mathcal{E} = 2 \,V$$. Calculate work done by the battery if the polarity of battery is reversed.

A $$6\, \mu J$$

B $$4\, \mu j$$

C $$3\, \mu J$$

D $$24\, \mu J$$

×

Initially, the capacitor is uncharged,

Qi = 0 When capacitor is connected to battery, the capacitor gets fully charge up to maximum value (Qmax).

Qf = Qmax = $$C\mathcal{E}$$

Qf = 3 × 10-6 × 2

Qf = $$6\, \mu C$$ Hence, the charge flow through battery is

$$\Delta$$Q = Qf – Qi

or,   $$\Delta$$Q = Qmax – 0

or,   $$\Delta$$Q = Qmax = $$C\mathcal{E}$$

$$\Delta$$Q = 3 × 10-6 × 2

$$\Delta$$$$Q = 6 \,\mu C$$ Work done by battery,

W1 = $$\Delta Q × \mathcal{E} =C\mathcal{E} ^{2}$$

W1 = 3 × 10-6 × (2)2

W1 = $$12\, \mu J$$ Now, the the polarity of battery is reversed. Before reversing the polarity of battery, charge on negative plate is – Q is,

Q'i = – $$6 \mu c$$  After reversing the battery, charge on negative plate is $$+ 6 \,\mu C$$.  Final charge on capacitor,

Q'= + Q = + $$C\mathcal{E}$$

Q'f = 3 × 10-6 × 2

Q'f = $$6 \,\mu C$$ Hence, charge flow through battery is

$$\Delta$$Q' = Q'f – Q'i

or,     $$\Delta$$Q' = + Q – (– Q)

or,     $$\Delta$$Q' = 2 Q

⇒ $$\Delta$$Q' = $$2 × 6 \mu C$$

⇒ $$\Delta$$Q' $$12 \mu C$$ Hence, Work done by battery,

W2 = $$\Delta$$Q' × $$\mathcal{E}$$= 2 Q × $$\mathcal{E}$$

=  $$2 C\mathcal{E}^{2}$$

W2 = 2 × 3 × 10-6 × (2)2

W2 = $$24 \,\mu J$$ ### An uncharged capacitor having capacitance $$C = 3 \,\mu F$$, is connected to a battery $$\mathcal{E} = 2 \,V$$. Calculate work done by the battery if the polarity of battery is reversed. A

$$6\, \mu J$$

.

B

$$4\, \mu j$$

C

$$3\, \mu J$$

D

$$24\, \mu J$$

Option D is Correct

# Potential Energy across a Capacitor

• Consider an uncharged capacitor C, connected to a battery with e.m.f. $$\mathcal{E}$$, as shown in figure.  • Potential energy of the charged capacitor is

$$U= \dfrac {1}{2} C\mathcal{E}^{2}$$

• Now, this charged capacitor is connected with another battery of same e.m.f. but polarity is reversed.  •  The initial potential energy when switch is open

$$U_i = \dfrac {1}{2} C\mathcal{E}^{2}$$

• The final potential energy when switch is closed

$$U_f = \dfrac {1}{2} C\mathcal{E}^{2}$$  • Hence, change in potential energy

$$\Delta$$U = Uf – Ui

or,  $$\Delta U = \dfrac{1}{2} C\mathcal{E}^{2} - \dfrac{1}{2}C\mathcal{E}^{2}$$

or, $$\Delta U = 0$$

Note :- The energy stored in the capacitor remains same even if the polarity of the battery is reversed.

Thus, work done by battery appears as heat.

#### An uncharged capacitor of capacitance $$C =4 \,\mu F$$ is connected with battery $$\mathcal{E} = 3 \,V$$. Calculate the change in potential energy when the polarity of battery is reversed.

A 18 J

B Zero

C – 18 J

D 36 J

×

Initial potential energy when battery is connected as shown in figure.

$$U_i = \dfrac {1}{2} C\mathcal{E}^{2}$$

⇒  $$U_i =\dfrac{1}{2}× 4 × 10^{-6}× (3)^{2}$$

⇒  $$U_i=18\, \mu J$$ Final potential energy when battery is reversed,

$$U_f = \dfrac {1}{2} C\mathcal{E}^{2}$$

⇒  $$U_f =\dfrac{1}{2}× 4 × 10^{-6}× (3)^{2}$$

⇒ $$U_f = 18\, \mu J$$ So, change in potential energy is given as

$$\Delta U = U_f - U_i$$

$$\Delta U$$ = $$18 \,\mu J$$ – $$18\, \mu J$$

$$\Delta U=0$$

### An uncharged capacitor of capacitance $$C =4 \,\mu F$$ is connected with battery $$\mathcal{E} = 3 \,V$$. Calculate the change in potential energy when the polarity of battery is reversed.

A

18 J

.

B

Zero

C

– 18 J

D

36 J

Option B is Correct

# Total Energy Stored in a Circuit

• Consider two capacitors having capacitance C1 and C2 are connected across a battery of e.m.f. $$\mathcal{E}$$, as shown in figure.  • Energy stored in capacitor C1

$$U_1 = \dfrac {1}{2} C_1\mathcal{E} ^{2}$$

• Energy stored in capacitor C2

$$U_2 = \dfrac {1}{2} C_2 \mathcal{E} ^{2}$$

• Total energy of the circuit

$$U = U_1 + U_2$$

$$\Rightarrow U = \dfrac{1}{2} C_1 \mathcal{E}^{2} + \dfrac {1}{2} C_2 \mathcal{E}^{2}$$

$$\Rightarrow U = \dfrac{1}{2}\mathcal{E} ^{2} ( C_1 + C_2)$$

#### Two capacitors of capacitance $$C_1=3\,\mu F$$  and  $$C_2=4\,\mu F$$  are connected in parallel across the battery of emf $$\mathcal{E}=2\,V$$, as shown in figure. Calculate the energy stored in each capacitor.

A $$3\,\mu J,\,4\,\mu J$$

B $$2\,\mu J,\,6\,\mu J$$

C $$6\,\mu J,\,8\,\mu J$$

D $$6\,\mu J,\,3\,\mu J$$

×

Energy stored in capacitor C1

$$U_1=\dfrac{1}{2}C_1 \mathcal{E}^2$$

Or,  $$U_1=\dfrac{1}{2}×3×10^{-6}×(2)^2$$

Or,  $$U_1=6\,\mu J$$ Energy stored in capacitor C2

$$U_2=\dfrac{1}{2}C_2 \mathcal{E}^2$$

Or,  $$U_2=\dfrac{1}{2}×4×10^{-6}×(2)^2$$

Or,  $$U_2=8\,\mu J$$ ### Two capacitors of capacitance $$C_1=3\,\mu F$$  and  $$C_2=4\,\mu F$$  are connected in parallel across the battery of emf $$\mathcal{E}=2\,V$$, as shown in figure. Calculate the energy stored in each capacitor. A

$$3\,\mu J,\,4\,\mu J$$

.

B

$$2\,\mu J,\,6\,\mu J$$

C

$$6\,\mu J,\,8\,\mu J$$

D

$$6\,\mu J,\,3\,\mu J$$

Option C is Correct

# Work Done by Battery

• Consider a capacitor of capacitance C, having an initial charge Qi .
• When it is connected with battery of emf $$\mathcal{E}$$, the capacitor gets charged up to a final value Qf .  • Initially till the switch is open, the charge on positive plate is Qi .
• When switch (S) is closed, the capacitor gets charged up to Qf

$$Q_f=C\mathcal{E}$$

• Hence, work done by battery

$$W=\Delta Q.\mathcal{E}$$

or, $$W=(Q_f-Q_i)\mathcal{E}$$

or, $$W=(C\mathcal{E}-Q_i)\mathcal{E}$$

#### A capacitor of capacitance $$C=2\,\mu F$$, connected with a battery of e.m.f $$\mathcal{E}=2\,V$$, as shown in figure. Calculate the work done by battery when the initial charge present on capacitor before connecting it to battery is $$Q_i=6\,\mu C$$.

A $$–2\,\mu J$$

B $$–6\,\mu J$$

C $$–3\,\mu J$$

D $$–4\,\mu J$$

×

The initial charge on capacitor before connecting to battery i.e, switch is open is,

$$Q_i=6\,\mu C$$ The maximum charge on capacitor after the battery is connected, i.e, switch is closed.

$$Q_f = C\mathcal{E}$$

or,    $$Q_f = 2 × 10^{-6} × 2$$

or,    $$Q_f =4\,\mu C$$ Hence, initial charge is greater than the maximum charge acquired by capacitor after connecting it to battery.

Qmax < Qi

Amount of charge supplied by battery,

$$\Delta Q = Q_f - Q_i$$

$$\Delta Q = 4 \,–\, 6 =\, – 2 \mu C$$ Hence, work done by battery

$$W = \Delta Q × \mathcal{E}$$

$$or, W = – 2 × 10^{-6} × 2$$

or,   $$W=\,–4\,\mu J$$

### A capacitor of capacitance $$C=2\,\mu F$$, connected with a battery of e.m.f $$\mathcal{E}=2\,V$$, as shown in figure. Calculate the work done by battery when the initial charge present on capacitor before connecting it to battery is $$Q_i=6\,\mu C$$. A

$$–2\,\mu J$$

.

B

$$–6\,\mu J$$

C

$$–3\,\mu J$$

D

$$–4\,\mu J$$

Option D is Correct

# Energy Radiated in the Form of Heat

• Consider a capacitor of capacitance C, connected to a battery with e.m.f. $$\mathcal{E}$$, as show in figure.
• Work done by battery is

$$W = (Q_f - Q_i) × \mathcal{E}$$  •  If initially, capacitor is uncharged,

$$Q_i = 0$$

so, Qf = Qmax =  $$C \mathcal{E}$$

• Hence, work done is

W = Qmax × $$\mathcal{E}$$

or,     W = $$C \mathcal{E} × \mathcal{E}$$

or,     W = $$C\mathcal{E}^{2}$$

• Also, energy stored in capacitor is

$$U = \dfrac{1}{2} C\mathcal{E}^{2}$$

Conclusion :- The amount of energy stored in capacitor is half of work done by battery. It means half of the energy is radiated as heat.

Heat = work done by battery - energy stored in capacitor

⇒ Heat = $$W -\dfrac{1}{2} C\mathcal{E}^{2}$$

⇒ Heat = $$C\mathcal{E}^{2} - \dfrac{1}{2}C\mathcal{E}^{2}$$

⇒ Heat = $$\dfrac{1}{2}C\mathcal{E}^{2}$$

#### An uncharged capacitor having capacitance $$C = 5 \,\mu F$$, is connected with a battery of e.m.f. $$\mathcal{E} = 10 \,V$$. Calculate the heat produced by battery.

A $$200\, \mu J$$

B $$250\, \mu J$$

C $$300\, \mu J$$

D $$350\, \mu J$$

×

Heat = work done by battery - energy stored in capacitor

⇒ Heat = $$W- \dfrac{1}{2} C\mathcal{E}^{2}$$

⇒ Heat = $$C\mathcal{E}^{2} - \dfrac{1}{2}C\mathcal{E}^{2}$$

⇒ Heat = $$\dfrac{1}{2}C\mathcal{E}^{2}$$

Heat = $$\dfrac {1}{2} × 5 × 10^{-6} × (10)^{2}$$

Heat = $$250× 10^{-6}$$

Heat = $$250\, \mu J$$

### An uncharged capacitor having capacitance $$C = 5 \,\mu F$$, is connected with a battery of e.m.f. $$\mathcal{E} = 10 \,V$$. Calculate the heat produced by battery.

A

$$200\, \mu J$$

.

B

$$250\, \mu J$$

C

$$300\, \mu J$$

D

$$350\, \mu J$$

Option B is Correct

# Work Done by External Force

• Consider an uncharged capacitor, as shown in figure.

Charge on capacitor Q = 0  • When switch is closed, the capacitor gets fully charged up to Q amount.

Charge on capacitor Q = $$C\mathcal{E}$$  • Assume that the external force is applied on plates of capacitor to increase the separation between them from d to 2d.  • Then capacitance of capacitor ,$$C^{'} = \dfrac{\mathcal{E} A}{d^{'}}$$

When d increase to 2d

$$C^{'} = \dfrac{\mathcal{E} A}{2 d} = \dfrac {C}{2}$$

• Hence, charge on capacitor will be

$$Q^{'} = C^{'} V$$

$$Q^{'} = \dfrac {C}{2} V = \dfrac {Q}{2}$$

• Initial potential energy

$$U_i = \dfrac {1}{2} C\mathcal{E} ^{2}$$

• Final potential energy

$$U_f = \dfrac{1}{2} C^{'}\mathcal{E}^{2}$$

⇒ $$U_f = \dfrac{1}{2}(\dfrac{C}{2})\mathcal{E} ^{2}$$

⇒ $$U_f = \dfrac{1}{4}\mathcal{E} ^{2}$$

• Work done in increasing the separation from d to 2d,

Wb = $$\Delta Q .\mathcal{E}$$

⇒ W=$$( Q_f - Q_i ) \mathcal{E}$$

⇒ W$$(\dfrac {Q}{2} - Q ) \mathcal{E}$$

⇒ W $$\dfrac{-Q}{2}\mathcal{E}$$

⇒ Wb = $$\dfrac {-1}{2}C\mathcal{E} ^{2}$$

• Hence, work done by external force can be determined by,

Ui + Wb = Uf + Wext

⇒ $$\dfrac {1}{2} C\mathcal{E} ^{2} - \dfrac {1}{2} C\mathcal{E} ^{2} = \dfrac {1}{4}C\mathcal{E}^{2}$$ + Wext

⇒ Wext $$\dfrac {-1}{4} C\mathcal{E} ^{2}$$

Note :- While moving the plates of capacitor slowly, heat is zero.

Heat = 0

#### A charged capacitor (C) is connected to a battery with e.m.f. $$\mathcal{E}$$. Chose the incorrect option when the separation between capacitor plate is halved.

A Wext = $$\dfrac {1}{2}C\mathcal{E}^{2}$$

B WB = $$\dfrac {1}{2}C\mathcal{E} ^{2}$$

C $$\Delta U = \dfrac {1}{2}C\mathcal{E}^{2}$$

D WB = $$C\mathcal{E}^{2}$$

×

When the separation between plates is halved

d' = $$d/2$$

$$C^{'}= \dfrac {\varepsilon _0A}{d^{'}} = \dfrac {{\varepsilon _0A}}{(d/2)}$$

$$C^{'} = \dfrac {2 \varepsilon _0A}{d} = 2C$$

$$C^{'} = 2C$$

Hence, capacitance is doubled.

Charge on capacitor plates (Q')

Q' = C' V

Q' = 2 CV

Q' = 2Q

Initial potential energy (Ui)

$$U_i = \dfrac {1}{2} C\mathcal{E} ^{2}$$

Final potential energy (Uf)

$$U_f = \dfrac {1}{2}C^{'} \mathcal{E}^{2}$$

$$U_f = \dfrac {1}{2} (2C) \mathcal{E} ^{2} = C \mathcal{E} ^{2}$$

Work done ( W)

Wb = $$\Delta Q .\mathcal{E}$$

⇒ W=$$( Q_f - Q_i )\mathcal{E}$$

⇒ W$$(2 Q - Q) \mathcal{E}$$

⇒ W$$Q \mathcal{E}$$

⇒ W$$C\mathcal{E}^{2}$$

Work done by external force (Wext)

Ui +Wb = Uf +Wext

$$\dfrac{1}{2} C\mathcal{E}^{2} + C\mathcal{E}^{2} = C\mathcal{E}^{2}$$ + Wext

Wext $$\dfrac{1}{2} C\mathcal{E}^{2}$$

Change in potential energy $$(\Delta \;U )$$

$$\Delta U = U_f - U_i$$

$$= C\mathcal{E} ^{2} - \dfrac{1}{2} C\mathcal{E} ^{2}$$

$$\dfrac{1}{2} C\mathcal{E}^{2}$$

Hence, option ( B ) is incorrect.

### A charged capacitor (C) is connected to a battery with e.m.f. $$\mathcal{E}$$. Chose the incorrect option when the separation between capacitor plate is halved.

A

Wext = $$\dfrac {1}{2}C\mathcal{E}^{2}$$

.

B

WB = $$\dfrac {1}{2}C\mathcal{E} ^{2}$$

C

$$\Delta U = \dfrac {1}{2}C\mathcal{E}^{2}$$

D

WB = $$C\mathcal{E}^{2}$$

Option B is Correct