Energy Stored in a Capacitor

The Concept of Stored Energy

Charging a capacitor involves transferring charges from one plate to another. Since the plates already possess like charges, we must do work against the electrostatic repulsive forces. This work done is stored in the capacitor as Electrostatic Potential Energy ($U$). This energy 'lives' in the electric field between the plates.

Energy Stored in a Capacitor

1. The Setup: Consider a capacitor of capacitance $C$ being charged. Let at some intermediate stage, the charge on the capacitor be $q$ and the potential difference be $v$. Then, $v = q/C$.

2. Small Work Done ($dW$): Suppose a small additional charge $dq$ is transferred from the negative plate to the positive plate. The work done $dW$ is: $$dW = v \cdot dq = \frac{q}{C} dq$$

3. Total Work Done ($W$): To find the total work done in charging the capacitor from an initial charge of 0 to a final charge $Q$, we integrate: $$W = \int_{0}^{Q} \frac{q}{C} dq$$ $$W = \frac{1}{C} \int_{0}^{Q} q \, dq = \frac{1}{C} \left[ \frac{q^2}{2} \right]_{0}^{Q}$$ $$W = \frac{Q^2}{2C}$$

4. Alternative Forms: Since $Q = CV$, we can write this energy $U$ in three equivalent ways:

1. Fundamental: $U = \frac{Q^2}{2C}$
2. Voltage-based: $U = \frac{1}{2} CV^2$
3. Charge-Voltage: $U = \frac{1}{2} QV$

Energy Density ($u$)

Energy density is defined as the energy stored per unit volume of the capacitor. For a parallel plate capacitor:

1. Energy: $U = \frac{1}{2} CV^2 = \frac{1}{2} \left( \frac{\epsilon_0 A}{d} \right) (Ed)^2$
2. Volume: $Volume = A \cdot d$
3. Density Calculation: $$u = \frac{U}{A \cdot d} = \frac{\frac{1}{2} \epsilon_0 A E^2 d}{A \cdot d}$$ $$u = \frac{1}{2} \epsilon_0 E^2$$

Note: This formula is universal. Any electric field $E$ in vacuum carries an energy density of $\frac{1}{2} \epsilon_0 E^2$.

🧠 Memory Capsule

• The Half Factor: Just like kinetic energy ($1/2 mv^2$), capacitor energy is $1/2 CV^2$.
• Work by Battery: If a battery of voltage $V$ charges a capacitor to charge $Q$, the battery does work $W_{bat} = QV$. However, only $1/2 QV$ is stored. The other 50% is lost as heat in the wires.
• Energy Density: Depends only on the strength of the electric field ($E^2$).

Example 1: Basic Energy Calculation

A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?

Solution:

1. Identify values: $C = 12 \times 10^{-12}$ F, $V = 50$ V.
2. Formula: $U = \frac{1}{2} CV^2$.
3. Step-by-step Calculation:
   • $U = 0.5 \times (12 \times 10^{-12}) \times (50)^2$
   • $U = 6 \times 10^{-12} \times 2500$
   • $U = 1.5 \times 10^{-8}$ J.
4. Final Answer: $1.5 \times 10^{-8}$ J.

Example 2: Energy from Charge

A capacitor is charged by a 200 V supply and stores 2 mC of charge. Find the energy stored.

Solution:

1. Identify values: $V = 200$ V, $Q = 2 \times 10^{-3}$ C.
2. Formula: $U = \frac{1}{2} QV$.
3. Calculation:
   • $U = 0.5 \times (2 \times 10^{-3}) \times 200$
   • $U = 10^{-3} \times 200 = 0.2$ J.
4. Final Answer: 0.2 J.

Example 3: Energy Density Calculation

A parallel plate capacitor has an electric field of $10^5$ V/m between its plates. Calculate the energy density.

Solution:

1. Identify values: $E = 10^5$ V/m, $\epsilon_0 = 8.854 \times 10^{-12}$.
2. Formula: $u = \frac{1}{2} \epsilon_0 E^2$.
3. Calculation:
   • $u = 0.5 \times 8.854 \times 10^{-12} \times (10^5)^2$
   • $u = 4.427 \times 10^{-12} \times 10^{10} = 4.427 \times 10^{-2}$ J/m$^3$.
4. Final Answer: $0.044$ J/m$^3$.

Example 4: Effect of Disconnecting Battery

A capacitor $C$ is charged to $V$ and the battery is disconnected. If the distance between plates is doubled, what is the new energy stored?

Solution:

1. Initial Energy: $U_i = \frac{Q^2}{2C}$.
2. Identify State: Battery disconnected means charge $Q$ is constant.
3. Capacitance Change: $d$ is doubled, so $C$ becomes $C/2$.
4. New Energy: $U_f = \frac{Q^2}{2(C/2)} = 2 \left( \frac{Q^2}{2C} \right) = 2 U_i$.
5. Final Answer: Energy doubles. (External work was done to pull the plates apart).

Example 5: Effect of Keeping Battery Connected

In Example 4, if the battery remains connected while the distance is doubled, what is the new energy?

Solution:

1. Initial Energy: $U_i = \frac{1}{2} CV^2$.
2. Identify State: Battery connected means potential $V$ is constant.
3. Capacitance Change: $C$ becomes $C/2$.
4. New Energy: $U_f = \frac{1}{2} (C/2) V^2 = \frac{1}{2} U_i$.
5. Final Answer: Energy is halved. (Charge flowed back into the battery).

Example 6: Energy Loss in Sharing

A 600 pF capacitor is charged to 200 V. It is then disconnected and connected to another uncharged 600 pF capacitor. Find the energy lost.

Solution:

1. Initial Energy: $U_i = \frac{1}{2} C_1 V_1^2 = 0.5 \times 600 \times 10^{-12} \times (200)^2 = 1.2 \times 10^{-5}$ J.
2. Common Potential ($V$): Since $C_1 = C_2$, $V = 200/2 = 100$ V.
3. Final Energy: $U_f = \frac{1}{2} (C_1 + C_2) V^2 = 0.5 \times (1200 \times 10^{-12}) \times 100^2 = 0.6 \times 10^{-5}$ J.
4. Energy Loss: $\Delta U = U_i - U_f = 0.6 \times 10^{-5}$ J.
5. Final Answer: $6 \times 10^{-6}$ J (or 6 $\mu$J).

Example 7: Work Done by Battery

A 10 $\mu$F capacitor is charged to 10 V. Calculate (a) the energy stored and (b) the work done by the battery.

Solution:

1. Energy Stored: $U = \frac{1}{2} CV^2 = 0.5 \times 10 \times 10^{-6} \times 10^2 = 5 \times 10^{-4}$ J.
2. Work by Battery: $W = QV = CV^2 = 10 \times 10^{-6} \times 10^2 = 10 \times 10^{-4}$ J.
3. Observation: Only half the battery's work is stored as energy.
4. Final Answer: (a) 0.5 mJ, (b) 1.0 mJ.

Example 8: Energy with Dielectric (Constant V)

A capacitor is connected to a battery. A dielectric $K = 3$ is inserted. How does the energy change?

Solution:

1. Constant: $V$ is constant.
2. Capacitance: $C' = 3C$.
3. Energy: $U' = \frac{1}{2} (3C) V^2 = 3 U$.
4. Final Answer: Energy triples.

Example 9: Energy with Dielectric (Constant Q)

A charged capacitor is isolated. A dielectric $K = 3$ is inserted. How does the energy change?

Solution:

1. Constant: $Q$ is constant.
2. Capacitance: $C' = 3C$.
3. Energy: $U' = \frac{Q^2}{2(3C)} = \frac{1}{3} U$.
4. Final Answer: Energy becomes one-third.

Example 10: Potential Energy of a Dipole (Review)

A dipole ($p = 10^{-29}$ C m) is held at $60^{\circ}$ in a field $E = 10^4$ N/C. Find its potential energy.

Solution:

1. Formula: $U = -pE \cos \theta$.
2. Calculation: $U = -(10^{-29}) \times (10^4) \times \cos 60^{\circ} = -10^{-25} \times 0.5 = -5 \times 10^{-26}$ J.
3. Final Answer: $-5 \times 10^{-26}$ J.