Energy Stored in Capacitor — Revision Notes

⚡ 30-Second Revision

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)

2-Minute Revision

The energy stored in a capacitor is electrical potential energy residing in its electric field, resulting from the work done to separate charges. The three primary formulas are $U = \frac{1}{2}CV^2$, $U = \frac{Q^2}{2C}$, and $U = \frac{1}{2}QV$.

Remember to use the appropriate formula based on whether voltage ($V$) or charge ($Q$) is constant in a given scenario. Energy density, the energy per unit volume, is given by $u = \frac{1}{2}epsilon E^2$.

A critical point for NEET is the effect of inserting a dielectric: if the battery remains connected, $V$ is constant, and energy increases ($U' = KU$); if the battery is disconnected, $Q$ is constant, and energy decreases ($U' = U/K$).

Also, when two charged capacitors are connected, total charge is conserved, but energy is lost as heat, calculated by $Delta U = \frac{1}{2} \frac{C_1C_2}{C_1 + C_2} (V_1 - V_2)^2$. Finally, the work done by the battery to charge a capacitor is $QV$, but only half of this ($rac{1}{2}QV$) is stored, with the other half dissipated as heat.

5-Minute Revision

Energy stored in a capacitor is the electrical potential energy accumulated in its electric field. This energy is a direct result of the work done by an external source to move charges against electrostatic forces.

The fundamental derivation starts from $dW = V dq = (q/C) dq$, leading to $U = int (q/C) dq = Q^2/(2C)$. Using $Q=CV$, this transforms into $U = \frac{1}{2}CV^2$ and $U = \frac{1}{2}QV$. These three forms are interchangeable, but choosing the right one simplifies problem-solving based on what quantities are constant.

For instance, if a capacitor is charged and then disconnected from the battery, its charge $Q$ remains constant, so $U = Q^2/(2C)$ is often preferred for subsequent calculations. If it remains connected to the battery, its voltage $V$ is constant, making $U = \frac{1}{2}CV^2$ more suitable.

The energy is actually stored in the electric field itself, not on the plates. The energy density $u$ (energy per unit volume) in an electric field $E$ is $u = \frac{1}{2}epsilon E^2$, where $epsilon$ is the permittivity of the medium. For vacuum, $epsilon = epsilon_0$. If a dielectric of constant $K$ is inserted, $epsilon = Kepsilon_0$.

Consider the impact of a dielectric: If the battery stays connected, $V$ is constant, $C$ becomes $KC$, and $U$ becomes $KU$. The battery does extra work. If the battery is disconnected, $Q$ is constant, $C$ becomes $KC$, and $U$ becomes $U/K$. The energy decrease is converted into mechanical work on the dielectric.

A common NEET problem involves connecting two charged capacitors. Total charge is conserved, but energy is lost due to resistance. The energy loss is $Delta U = \frac{1}{2} \frac{C_1C_2}{C_1 + C_2} (V_1 - V_2)^2$. Remember that only half the work done by the charging battery ($QV$) is stored as potential energy; the other half is dissipated as heat.

Prelims Revision Notes

1
Definition: — Energy stored in a capacitor is electrical potential energy in its electric field due to charge separation.
2
Formulas for Stored Energy (U):

* $U = \frac{1}{2}CV^2$ (Most common, use when V is constant) * $U = \frac{Q^2}{2C}$ (Use when Q is constant) * $U = \frac{1}{2}QV$

1
Units: — Energy in Joules (J), Capacitance in Farads (F), Voltage in Volts (V), Charge in Coulombs (C).
2
Energy Density (u): — Energy per unit volume.

* In vacuum: $u = \frac{1}{2}\epsilon_0 E^2$ * In a dielectric medium (dielectric constant $K$): $u = \frac{1}{2}K\epsilon_0 E^2 = \frac{1}{2}\epsilon E^2$

1
Effect of Dielectric (Dielectric Constant K):

* Battery Connected (V constant): * Capacitance: $C' = KC$ * Charge: $Q' = KQ$ * Energy: $U' = KU$ (Energy increases, battery does more work) * Battery Disconnected (Q constant): * Capacitance: $C' = KC$ * Voltage: $V' = V/K$ * Energy: $U' = U/K$ (Energy decreases, converted to mechanical work)

1
Energy Loss on Connecting Two Charged Capacitors:

* When $C_1$ (charged to $V_1$) and $C_2$ (charged to $V_2$) are connected in parallel (like polarities). * Total charge is conserved: $Q_{total} = C_1V_1 + C_2V_2$ * Common potential: $V_{common} = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}$ * Energy Loss: $\Delta U = U_{initial} - U_{final} = \frac{1}{2} \frac{C_1C_2}{C_1 + C_2} (V_1 - V_2)^2$ * Energy is always lost (dissipated as heat) unless $V_1 = V_2$.

1
Work Done by Battery vs. Stored Energy:

* Work done by battery: $W_B = QV$ * Energy stored: $U = \frac{1}{2}QV$ * Relationship: $W_B = 2U$. Half the work done by the battery is dissipated as heat during charging.

Vyyuha Quick Recall

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.