Why is only half the work done by the battery stored as energy in a capacitor?

When a capacitor is charged by a battery, the battery does work $W_{battery} = QV$, where $Q$ is the total charge transferred and $V$ is the final potential difference across the capacitor (equal to the battery voltage). However, the energy stored in the capacitor is $U = rac{1}{2}QV$. The difference, $W_{battery} - U = QV - rac{1}{2}QV = rac{1}{2}QV$, is dissipated as heat in the connecting wires and the internal resistance of the battery during the charging process. This energy loss occurs because the charging current flows through a resistive path, and energy is lost as $I^2Rt$ heat.

What is energy density in the context of a capacitor?

Energy density refers to the amount of energy stored per unit volume within the electric field of the capacitor. For a parallel plate capacitor in vacuum, it is given by $u = rac{1}{2}epsilon_0 E^2$, where $epsilon_0$ is the permittivity of free space and $E$ is the magnitude of the electric field between the plates. This concept is important because it highlights that the energy is not stored 'on the plates' but rather 'in the space' occupied by the electric field.

How does inserting a dielectric affect the stored energy if the capacitor remains connected to the battery?

If the capacitor remains connected to the battery, the potential difference $V$ across its plates stays constant. When a dielectric with constant $K$ is inserted, the capacitance increases to $C' = KC$. Since $U = rac{1}{2}CV^2$, the new stored energy $U' = rac{1}{2}C'V^2 = rac{1}{2}(KC)V^2 = K(rac{1}{2}CV^2) = KU$. Thus, the stored energy increases by a factor of $K$. This additional energy comes from the battery, which does more work to supply extra charge to the capacitor.

How does inserting a dielectric affect the stored energy if the battery is disconnected?

If the battery is disconnected before inserting the dielectric, the charge $Q$ on the capacitor plates remains constant (as there's no external circuit for charge to flow). When a dielectric with constant $K$ is inserted, the capacitance increases to $C' = KC$. Using the formula $U = rac{Q^2}{2C}$, the new stored energy $U' = rac{Q^2}{2C'} = rac{Q^2}{2(KC)} = rac{1}{K}(rac{Q^2}{2C}) = rac{U}{K}$. Thus, the stored energy decreases by a factor of $K$. This decrease in energy is converted into mechanical work done by the electric field on the dielectric material as it is pulled into the capacitor.

Is energy conserved when two charged capacitors are connected together?

No, generally energy is not conserved when two charged capacitors are connected. While the total charge on the isolated system of capacitors is conserved, there is typically a loss of energy. This energy loss occurs primarily as heat dissipated in the connecting wires due to their resistance, as charge flows to equalize the potential difference. The energy loss is given by $Delta U = rac{1}{2} rac{C_1C_2}{C_1 + C_2} (V_1 - V_2)^2$, which is always positive unless $V_1 = V_2$ (no charge flow).

Can a capacitor store an infinite amount of energy?

No, a capacitor cannot store an infinite amount of energy. The amount of energy it can store is limited by its capacitance and the maximum voltage it can withstand before dielectric breakdown occurs. Every capacitor has a maximum operating voltage, beyond which the dielectric material between its plates will break down, leading to a short circuit and permanent damage. This breakdown voltage limits the maximum charge and thus the maximum energy that can be stored.

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Energy Stored in Capacitor

Physics

NEET UG

Version 1Updated 22 Mar 2026

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Definition Deep Explanation Core Principles NEET Importance Problem Strategy Practice Problems MCQ Practice Quick Revision

The energy stored in a capacitor represents the electrical potential energy accumulated within its electric field when it is charged. This energy is a direct consequence of the work done by an external source (like a battery) to separate positive and negative charges and store them on the capacitor plates against the repulsive forces between like charges and attractive forces between unlike charge…

Quick Summary

The energy stored in a capacitor is the electrical potential energy accumulated in its electric field when it is charged. This energy originates from the work done by an external source, like a battery, to separate charges and place them on the capacitor plates.

As charge accumulates, the potential difference across the plates increases, requiring more work to transfer additional charge. The total work done is stored as potential energy. The fundamental formulas for this stored energy are $U = \frac{1}{2}CV^2$ , $U = \frac{Q^2}{2C}$ , and $U = \frac{1}{2}QV$ , where $C$ is capacitance, $V$ is voltage, and $Q$ is charge.

The energy is actually stored in the electric field itself, with an energy density of $u = \frac{1}{2}epsilon E^2$ . When a dielectric is introduced, the stored energy changes: it increases if the capacitor remains connected to the battery (constant $V$ ), and it decreases if the battery is disconnected (constant $Q$ ).

A crucial point for NEET is that only half the work done by the battery is stored as energy, with the other half dissipated as heat during charging. Also, when charged capacitors are connected, total charge is conserved, but energy is typically lost due to resistance.

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Key Concepts

Derivation of Stored Energy (

U = \frac{Q^2}{2C}

)

The energy stored in a capacitor is the total work done to charge it. Imagine transferring infinitesimally…

Energy Density in Electric Field

The energy stored in a capacitor is physically located in the electric field between its plates. For a…

Energy Loss During Charge Redistribution

When two charged capacitors are connected, charge flows until a common potential is reached. This process is…

Energy Stored (U):

- $U = \frac{1}{2}CV^2$ - $U = \frac{Q^2}{2C}$ - $U = \frac{1}{2}QV$

Energy Density (u):

- In vacuum: $u = \frac{1}{2}\epsilon_0 E^2$ - In dielectric: $u = \frac{1}{2}K\epsilon_0 E^2 = \frac{1}{2}\epsilon E^2$

Effect of Dielectric (K):

- Battery Connected (V constant): $C' = KC$ , $U' = KU$ - Battery Disconnected (Q constant): $C' = KC$ , $U' = U/K$

Energy Loss (Connecting Capacitors):

- $\Delta U = \frac{1}{2} \frac{C_1C_2}{C_1 + C_2} (V_1 - V_2)^2$

**Work by Battery (

W_B

):**

- $W_B = QV = 2U$ (Half energy stored, half dissipated)

To remember the energy formulas and dielectric effects:

'C-V-Q, Half-Squared-Over-Two'

Capacitance, Voltage, Quantity (Charge)

Half —

CV^2

Squared —

Q^2

Over Two

C

And Half

QV

'Dielectric Dilemma: Connected V, Disconnected Q'

Connected — to battery: Voltage is constant. Energy increases (

U \to KU

Disconnected — from battery: Quantity (Charge) is constant. Energy decreases (

U \to U/K

This helps remember which quantity stays constant and how energy changes in each scenario.

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