Why does the dielectric material in a capacitor heat up when the capacitor is charged and discharged repeatedly?

The energy utilized during the polarization of a dielectric is not fully recovered during depolarization. Some energy is adrift during the charging and discharging processes of the capacitors, and this lost energy manifests as heat.

Why is there energy loss when two charged conductors at different potentials are connected by a conducting wire?

As charge flows through the connecting wires, some of the potential energy of the charges is converted into heat energy, which is then dissipated into the surroundings.

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JEE Physics

Energy Stored in a Capacitor

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Energy Stored In A Capacitor

Capacitors are essential elements in electrical and electronic circuits, crucial for energy storage and management. When a voltage is applied across a capacitor, it accumulates electrical energy in the electric field formed between its plates. This stored energy can be discharged as needed, which makes capacitors indispensable for a wide range of applications, including stabilizing voltage in power supplies and operating timing circuits.

1.0Definition Energy Stored In a Capacitor

A capacitor is a device designed to store electrical energy. The process of charging a capacitor entails transferring electric charges from one plate to another. The work done during this charging process is stored as electrical potential energy within the capacitor.
This energy is provided by the battery, utilizing its stored chemical energy, and can be recovered by discharging the capacitors.

2.0Expression For Energy Stored In a Capacitor

Consider a capacitor of C Capacitance, initially both the plates are uncharged, suppose positive charge is placed to plate 2 to plate 1,in this process external work is done.

$V = \frac{W}{q} = \frac{d W}{d q}$

$d W = V d q = \frac{q}{C} d q$

$W = \int d W = \int_{0}^{Q} \frac{q}{C} d q = \frac{1}{C} \frac{Q ^{2}}{2} = \frac{Q ^{2}}{2 C}$

$U = \frac{Q ^{2}}{2 C} = \frac{1}{2} C V^{2} = \frac{1}{2} Q V$

Energy Stored In Series Combinations of Capacitors

$U = \frac{Q ^{2}}{2 C _{1}} + \frac{Q ^{2}}{2 C _{2}} + \frac{Q ^{2}}{2 C _{3}} + \dots$ .

Energy Stored In Parallel Combinations of Capacitors

$U = \frac{1}{2} C_{1} V^{2} + \frac{1}{2} C_{2} V^{2} + \frac{1}{2} C_{3} V^{2} + \dots\dots\dots\dots\dots ....$

3.0Energy Density For Parallel Plate Capacitor

In Parallel Plate Capacitor, potential energy stored in the form of electric field i.e. in the space between two plates and volume of this space is (A × d).

$Energy Density = \frac{Energy}{Volume} = \frac{\frac{1}{2} C V ^{2}}{A d}$

$C = \frac{ϵ _{0} A}{d}, V = E d$

$Energy Density = \frac{Energy}{Volume} = \frac{\frac{1 ϵ _{A} A}{2} ( E d ) ^{2}}{A d} = \frac{1}{2} ϵ_{0} E^{2}$

$E = \frac{σ}{ϵ _{0}}$

$Energy Density = \frac{1}{2} \frac{σ ^{2}}{ϵ _{0}}$

4.0Charging Of Parallel Plate Capacitor By Battery

When a capacitor is charged by a battery then the battery charges it till then its potential difference becomes equal to EMF of the battery.
Whenever any capacitor (Initially charged or uncharged) is connected to battery then its final voltage is always equal to emf of battery.
Final Potential Energy of Capacitor = $\frac{1}{2} C V^{2}$
Work Done By Battery,

$W = Charge supplied by battery \times EMF$

W=q V

$Energy Supplied by battery (W) = C V^{2}$

From conservation of Energy,

$W_{battery} = Δ U + Heat Loss$

$C V^{2} = (\frac{1}{2} C V^{2} - 0) + Heat Loss$

$Heat Loss = \frac{1}{2} C V^{2}$

When an uncharged capacitor is associated with a battery then 50% of energy delivered by the battery is stored in the capacitor and the remaining 50% will be lost.
Energy loss does not depend on the resistance of the circuit.

Note: When initially capacitor is charged then heat loss is not equal to $\frac{1}{2} C V^{2}$ , find heat loss by use of following concept

W_{by battery} = U + Heat loss

Potential Energy of Conducting Sphere

$U = \frac{K Q ^{2}}{2 R} = \frac{1}{2} \frac{Q ^{2}}{4 π ϵ _{0} r}$

5.0Effect of Dielectric On Energy Stored

Case 1.When the battery is kept disconnected from the capacitor

$U = \frac{1}{2} C V^{2} = \frac{1}{2} (k C_{0}) (\frac{V _{0}}{k})^{2} = \frac{1}{k} \cdot \frac{1}{2} C_{0} V_{0}^{2} = \frac{U _{0}}{k}$ , Energy stored decreased by k

Case 2.When the battery remains connected across the capacitor

$U = \frac{1}{2} C V^{2} = \frac{1}{2} (k C_{0}) (V_{0})^{2} = k \cdot \frac{1}{2} C_{0} V_{0}^{2} = k U_{0^{'}}$ Energy stored in the capacitor is increased k times.

Work Done By External Agent to Charge A Conductor

Work done by an external agent to bring a small charged element dq from infinity to surface of conductor.

$d W = V d q = \frac{q}{c} d q$

Work done by external agent to charge the conductor from q₁ to q₂ is

$W = \int_{q_{1}}^{q_{2}} \frac{q}{c} \cdot d q \Rightarrow W = \frac{1}{C} (\frac{q ^{2}}{2})_{q_{1}}^{q_{2}}$

$q_{1} = C V_{1}, q_{2} = C V_{2}$

Work done by external agent to increase the potential of conductor from V₁ to V₂

$W = \frac{1}{2} C (V_{2}^{2} - V_{1}^{2})$

Example: Initially an uncharged capacitor having capacitance C is united across a battery of emf V. Now the capacitor is disconnected and then reconnected across the same battery but with reversed polarity. Find heat loss in this process.

Solution: Total Charge Flow $Q_{f} - Q_{i} = (C V) - (- C V) = 2 C V$

Work done by battery = $(2 C V) (V) = 2 C V^{2}$

$Heat loss = W.D by battery - (Change in stored energy)$

$Heat Loss = 2 C V^{2} - (\frac{1}{2} C V^{2} - \frac{1}{2} C V^{2}) = 2 C V^{2}$

6.0Sample Questions on Energy Stored In a Capacitor

Q-1.How can you connect two capacitors across a battery—either in series or in parallel—to maximize the total charge and total energy stored?

Solution:

$Total Charge, q = C V$

$Total Energy, U = \frac{1}{2} C V^{2}$

As V is constant and $C_{P} > C_{S}$ , so the capacitors must be connected in parallel for stronger charge and energy.

Q-2.What occurs to the energy stored in a capacitor if the plates of a charged parallel plate capacitor are pulled apart while the battery remains connected?

Solution: When the plates are pulled apart, the capacitance decreases. Since the battery remains connected, the potential difference stays constant. As a result, the energy stored in the capacitor decreases. $(U = \frac{1}{2} C V^{2})$

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Energy Stored In A Capacitor

Capacitors are essential elements in electrical and electronic circuits, crucial for energy storage and management. When a voltage is applied across a capacitor, it accumulates electrical energy in the electric field formed between its plates. This stored energy can be discharged as needed, which makes capacitors indispensable for a wide range of applications, including stabilizing voltage in power supplies and operating timing circuits.

1.0Definition Energy Stored In a Capacitor

A capacitor is a device designed to store electrical energy. The process of charging a capacitor entails transferring electric charges from one plate to another. The work done during this charging process is stored as electrical potential energy within the capacitor.
This energy is provided by the battery, utilizing its stored chemical energy, and can be recovered by discharging the capacitors.

2.0Expression For Energy Stored In a Capacitor

Consider a capacitor of C Capacitance, initially both the plates are uncharged, suppose positive charge is placed to plate 2 to plate 1,in this process external work is done.

$V = \frac{W}{q} = \frac{d W}{d q}$

$d W = V d q = \frac{q}{C} d q$

$W = \int d W = \int_{0}^{Q} \frac{q}{C} d q = \frac{1}{C} \frac{Q ^{2}}{2} = \frac{Q ^{2}}{2 C}$

$U = \frac{Q ^{2}}{2 C} = \frac{1}{2} C V^{2} = \frac{1}{2} Q V$

Energy Stored In Series Combinations of Capacitors

$U = \frac{Q ^{2}}{2 C _{1}} + \frac{Q ^{2}}{2 C _{2}} + \frac{Q ^{2}}{2 C _{3}} + \dots$ .

Energy Stored In Parallel Combinations of Capacitors

$U = \frac{1}{2} C_{1} V^{2} + \frac{1}{2} C_{2} V^{2} + \frac{1}{2} C_{3} V^{2} + \dots\dots\dots\dots\dots ....$

3.0Energy Density For Parallel Plate Capacitor

In Parallel Plate Capacitor, potential energy stored in the form of electric field i.e. in the space between two plates and volume of this space is (A × d).

$Energy Density = \frac{Energy}{Volume} = \frac{\frac{1}{2} C V ^{2}}{A d}$

$C = \frac{ϵ _{0} A}{d}, V = E d$

$Energy Density = \frac{Energy}{Volume} = \frac{\frac{1 ϵ _{A} A}{2} ( E d ) ^{2}}{A d} = \frac{1}{2} ϵ_{0} E^{2}$

$E = \frac{σ}{ϵ _{0}}$

$Energy Density = \frac{1}{2} \frac{σ ^{2}}{ϵ _{0}}$

4.0Charging Of Parallel Plate Capacitor By Battery

When a capacitor is charged by a battery then the battery charges it till then its potential difference becomes equal to EMF of the battery.
Whenever any capacitor (Initially charged or uncharged) is connected to battery then its final voltage is always equal to emf of battery.
Final Potential Energy of Capacitor = $\frac{1}{2} C V^{2}$
Work Done By Battery,

$W = Charge supplied by battery \times EMF$

W=q V

$Energy Supplied by battery (W) = C V^{2}$

From conservation of Energy,

$W_{battery} = Δ U + Heat Loss$

$C V^{2} = (\frac{1}{2} C V^{2} - 0) + Heat Loss$

$Heat Loss = \frac{1}{2} C V^{2}$

When an uncharged capacitor is associated with a battery then 50% of energy delivered by the battery is stored in the capacitor and the remaining 50% will be lost.
Energy loss does not depend on the resistance of the circuit.

Note: When initially capacitor is charged then heat loss is not equal to $\frac{1}{2} C V^{2}$ , find heat loss by use of following concept

W_{by battery} = U + Heat loss

Potential Energy of Conducting Sphere

$U = \frac{K Q ^{2}}{2 R} = \frac{1}{2} \frac{Q ^{2}}{4 π ϵ _{0} r}$

5.0Effect of Dielectric On Energy Stored

Case 1.When the battery is kept disconnected from the capacitor

$U = \frac{1}{2} C V^{2} = \frac{1}{2} (k C_{0}) (\frac{V _{0}}{k})^{2} = \frac{1}{k} \cdot \frac{1}{2} C_{0} V_{0}^{2} = \frac{U _{0}}{k}$ , Energy stored decreased by k

Case 2.When the battery remains connected across the capacitor

$U = \frac{1}{2} C V^{2} = \frac{1}{2} (k C_{0}) (V_{0})^{2} = k \cdot \frac{1}{2} C_{0} V_{0}^{2} = k U_{0^{'}}$ Energy stored in the capacitor is increased k times.

Work Done By External Agent to Charge A Conductor

Work done by an external agent to bring a small charged element dq from infinity to surface of conductor.

$d W = V d q = \frac{q}{c} d q$

Work done by external agent to charge the conductor from q₁ to q₂ is

$W = \int_{q_{1}}^{q_{2}} \frac{q}{c} \cdot d q \Rightarrow W = \frac{1}{C} (\frac{q ^{2}}{2})_{q_{1}}^{q_{2}}$

$q_{1} = C V_{1}, q_{2} = C V_{2}$

Work done by external agent to increase the potential of conductor from V₁ to V₂

$W = \frac{1}{2} C (V_{2}^{2} - V_{1}^{2})$

Example: Initially an uncharged capacitor having capacitance C is united across a battery of emf V. Now the capacitor is disconnected and then reconnected across the same battery but with reversed polarity. Find heat loss in this process.

Solution: Total Charge Flow $Q_{f} - Q_{i} = (C V) - (- C V) = 2 C V$

Work done by battery = $(2 C V) (V) = 2 C V^{2}$

$Heat loss = W.D by battery - (Change in stored energy)$

$Heat Loss = 2 C V^{2} - (\frac{1}{2} C V^{2} - \frac{1}{2} C V^{2}) = 2 C V^{2}$

6.0Sample Questions on Energy Stored In a Capacitor

Q-1.How can you connect two capacitors across a battery—either in series or in parallel—to maximize the total charge and total energy stored?

Solution:

$Total Charge, q = C V$

$Total Energy, U = \frac{1}{2} C V^{2}$

As V is constant and $C_{P} > C_{S}$ , so the capacitors must be connected in parallel for stronger charge and energy.

Q-2.What occurs to the energy stored in a capacitor if the plates of a charged parallel plate capacitor are pulled apart while the battery remains connected?

Solution: When the plates are pulled apart, the capacitance decreases. Since the battery remains connected, the potential difference stays constant. As a result, the energy stored in the capacitor decreases. $(U = \frac{1}{2} C V^{2})$

Frequently Asked Questions

Why does the dielectric material in a capacitor heat up when the capacitor is charged and discharged repeatedly?

Why is there energy loss when two charged conductors at different potentials are connected by a conducting wire?

Frequently Asked Questions

Why does the dielectric material in a capacitor heat up when the capacitor is charged and discharged repeatedly?

Why is there energy loss when two charged conductors at different potentials are connected by a conducting wire?

Join ALLEN!

Energy Stored In A Capacitor

1.0Definition Energy Stored In a Capacitor

2.0Expression For Energy Stored In a Capacitor

3.0Energy Density For Parallel Plate Capacitor

4.0Charging Of Parallel Plate Capacitor By Battery

Potential Energy of Conducting Sphere

5.0Effect of Dielectric On Energy Stored

Work Done By External Agent to Charge A Conductor

6.0Sample Questions on Energy Stored In a Capacitor

Table of Contents

Join ALLEN!

Energy Stored In A Capacitor

1.0Definition Energy Stored In a Capacitor

2.0Expression For Energy Stored In a Capacitor

3.0Energy Density For Parallel Plate Capacitor

4.0Charging Of Parallel Plate Capacitor By Battery

Potential Energy of Conducting Sphere

5.0Effect of Dielectric On Energy Stored

Work Done By External Agent to Charge A Conductor

6.0Sample Questions on Energy Stored In a Capacitor

Table of Contents