18.6 ENERGY STORED IN CHARGED CAPACITOR
18.6.1 Energy Stored in a Charged Capacitor
1. Formula
A capacitor, as we have seen above, is a system of two conductors with charge Q and –Q. To determine the energy stored in this configuration, consider initially two uncharged conductors 1 and 2. Imagine next a process of transferring charge from conductor 2 to conductor 1 bit by bit, so that at the end, conductor 1 gets charge Q. By charge conservation, conductor 2 has charge –Q at the end (Figure 18.20).
Figure 18.20: (a) Work done in a small step of building charge on conductor 1 from $Q′$ to $Q′ + \delta Q′$. (b) Total work done in charging the capacitor may be viewed as stored in the energy of electric field between the plates.
In transferring positive charge from conductor 2 to conductor 1, work will be done externally, since at any stage conductor 1 is at a higher potential than conductor 2. To calculate the total work done, we first calculate the work done in a small step involving transfer of an infinitesimal (i.e., vanishingly small) amount of charge. Consider the intermediate situation when the conductors 1 and 2 have charges $Q′$ and $–Q′$ respectively. At this stage, the potential difference V′ between conductors 1 to 2 is $Q′/C$, where $C$ is the capacitance of the system. Next imagine that a small charge $\delta Q′$ is transferred from conductor 2 to 1. Work done in this step ($\delta W$), resulting in charge $Q′$ on conductor 1 increasing to $Q′+ \delta Q′$, is given by
$\delta W=V'.\delta Q=\frac{Q'}{C}\delta Q' \quad (1)$
Since $\delta Q'$ can be made as small as we like, Eq. (1) can be written as
$\delta W=\frac{1}{2C}[(Q'+\delta Q')^2 -Q'^2] \quad (2)$
Equations (1) and (2) are identical because the term of second order in $\delta Q′$, i.e., $\delta Q′/2C$, is negligible, since $\delta Q′$ is arbitrarily small. The total work done (W) is the sum of the small work ($\delta W$) over the very large number of steps involved in building the charge Q′ from zero to Q.
$W=\Sigma_{sum \ over \ all \ steps} \delta W$
$= \Sigma_{sum \ over \ all \ steps} \frac{1}{2C}[(Q'+\delta Q')^2 -Q'^2] \quad (3)$
$=\frac{1}{2C}[(\delta Q'^2 -0)+(2\delta Q')^2-\delta Q'^2+(3\delta Q')^2-(2\delta Q')^2+...$
$+Q^2-(Q-\delta Q')^2]$
$W=\frac{1}{2C}[(\delta Q'^2 -0)=\frac{Q^2}{2C} \quad (4)$
The same result can be obtained directly from Eq. (1) by integration
$W=\int_{0}^{Q}\frac{Q'}{C}\delta Q'$
$W=\frac{1}{C}\frac{Q'^2}{2}\Big|_{0}^{Q}=\frac{Q^2}{2C}$
This is not surprising since integration is nothing but summation of a large number of small terms.
We can write the final result, Eq. (4) in different ways
$\boxed{W=\frac{Q^2}{2C}=\frac{1}{2}CV^2=\frac{1}{2}QV} \quad (5)$
Since electrostatic force is conservative, this work is stored in the form of potential energy of the system. For the same reason, the final result for potential energy [Eq. (5)] is independent of the manner in which the charge configuration of the capacitor is built up. When the capacitor discharges, this stored-up energy is released. It is possible to view the potential energy of the capacitor as ‘stored’ in the electric field between the plates. To see this, consider for simplicity, a parallel plate capacitor [of area A (of each plate) and separation d between the plates].
Energy stored in the capacitor
$W=\frac{Q^2}{2C}=\frac{(A\sigma)^2}{2}\times \frac{d}{\epsilon_0A} \quad (6)$
The surface charge density σ is related to the electric field E between the plates,
$E=\frac{\sigma}{\epsilon_0} \quad (7)$
From Eqs. (2.74) and (2.75) , we get
Energy stored in the capacitor
$u=\frac{1}{2}\epsilon_0E^2 \times Ad \quad (8)$
Note that Ad is the volume of the region between the plates (where electric field alone exists). If we define energy density as energy stored per unit volume of space, Eq (2.76) shows that
Energy density of electric field
$\boxed{u=\frac{1}{2}\epsilon_0E^2} \quad (9)$
Though we derived Eq. (9) for the case of a parallel plate capacitor, the result on energy density of an electric field is, in fact, very general and holds true for electric field due to any configuration of charges.
2. V-Q Graph
Figure 18.21 Relationship between potential difference (V) and charge (Q) in a capacitor. The linear graph indicates that voltage is directly proportional to charge. The shaded area under the V–Q graph represents the electric potential energy stored in the capacitor, which is equal to half the product of charge and voltage (U = ½QV)
A graph of the p.d. across a capacitor plotted against the charge stored by the capacitor will yield a straight line which passes through the origin, as shown in Figure 18.21. Suppose that the charge stored in the capacitor is Q when a p.d. V exist across capacitor. Then the electric energy W stored in the capacitor is given by
$W=\frac{1}{2}QV$
Referring to the Q - V graph, we notice that the area show the graph is given by $\frac{1}{2}QV$. Hence the area below the graph is a V - Q (or Q - V) graph for a capacitor represents the electric energy stored by the capacitor.
Example 18.8
(a) What is the magnitude of the electric field required to store 1.00 J of electric potential energy in a volume of $1.00 \ m^3$ in vacuum? (b) If the field magnitude is 10 times larger than that, how much energy is stored per cubic meter?
Answer
(a) The desired energy density is $u=1.00 \ J/m^3$. Then from Eq. (24.11),
$E=\sqrt{\frac{2u}{\epsilon_0}}$
$E=\sqrt{\frac{2(1.00 \ J/m^3)}{8.85 \times 10^{-12} \ C^2/N.m^2}}$
$E=4.75 \times 10^5 \ N/C$
(b) Equation (9) shows that u is proportional to $E^2$. If increases by a factor of 10, u increases by a factor of $10^2 =100$, so the energy density becomes $u=100 \ J/m^3$.
Example 18.9
Refer to the circuit shown in Figure 18.21. Determine
(a) the total capacitance of the circuit
(b) the electric energy stored in each capacitor
Answer
(a) $\frac{1}{C_T}=\frac{1}{30}+\frac{1}{60}=\frac{1}{20}$
$C_T=20 \ \mu F$
(b) charge stored in each capacitor = charge stored in the single equivalent capacitor of capacitance $C_T$
$=C_TV$
$=(20)(6)=120 \ \mu C$
For the 30 $\mu F$ capacitor,
$W_1=\frac{1}{2}\frac{Q^2}{C_1}$
$W_1=\frac{1}{2}\frac{(120 \ \mu C)^2}{2(30 \ \mu F)}=240 \ \mu J$
For the 60 $\mu F$ capacitor,
$W_1=\frac{1}{2}\frac{Q^2}{C_2}$
$W_1=\frac{1}{2}\frac{(120 \ \mu C)^2}{2(60 \ \mu F)}=120 \ \mu J$
Example 18.10
We connect a capacitor $C_1=8.0 \ \mu F$ to a power supply, charge it to a potential difference $V_0=120 \ V$, and disconnect the power supply (Figure. 18.22). Switch is open. (a) What is the charge $Q_0$ on $C_1$? (b) What is the energy stored in $C_1$? (c) Capacitor $C_2=4.0 \ \mu F$ is initially uncharged. We close switch S. After charge no longer flows, what is the potential difference across each capacitor, and what is the charge on each capacitor? (d) What is the final energy of the system?
Figure 18.22 When the switch is closed, the charged capacitor is connected to an uncharged capacitor The center part of the switch is an insulating handle; charge can flow only between the two upper terminals and between the two lower terminals.
Answer
In parts (a) and (b) we find the charge $Q_0$ and stored energy $U_{initial}$ for the single charged capacitor $C_1$ using Eqs. $C=\frac{Q}{V}$ and (24.9), respectively. After we close switch S, one wire connects the upper plates of the two capacitors and another wire connects the lower plates; the capacitors are now connected in parallel. In part (c) we use the character of the parallel connection to determine how $Q_0$ is shared between the two capacitors. In part (d) we again use Eq. (5) to find the energy stored in capacitors $C_1$ and $C_2$ the energy of the system is the sum of these values.
(a) The initial charge $Q_0$ on $C_1$ is
$Q_0=C_1V_0=(8.0 \ \mu F)(120 \ V)=960 \ \mu C$
(b) The energy initially stored in $C_1$ is
$U_{initial}=\frac{1}{2}Q_0V_0$
$U_{initial}=\frac{1}{2}(960 \times 10^{-6} \ C)(120 \ V)=0.058 \ J$
(c) When we close the switch, the positive charge $Q_0$ is distributed over the upper plates of both capacitors and the negative charge $-Q_0$ is distributed over the lower plates. Let $Q_1$ and $Q_2$ be the magnitudes of the final charges on the capacitors. Conservation of charge requires that $Q_1+Q_2=Q_0$. The potential difference between the plates is the same for both capacitors because they are connected in parallel, so the charges are $Q_1=C_1V$ and $Q_2=C_2V$. We now have three independent equations relating the three unknowns $Q_1$, $Q_2$, and V. Solving these, we find
$V=\frac{Q_0}{C_1+C_2}=\frac{960 \ \mu C}{8.0 \ \mu F+4.0 \ \mu F}=80 \ V$
$Q_1=640 \ \mu C$; $Q_2=320 \ \mu C$
(d) The final energy of the system is
$u_{final}=\frac{1}{2}Q_1V+\frac{1}{2}Q_2V=\frac{1}{2}Q_0V$
$u_{final}=\frac{1}{2}(960 \times 10^{-6} \ C)(80 \ V)=0.038 \ J$
Applications of Energy Stored in Capacitors
The energy stored in capacitors plays an important role in many electrical and electronic systems. Because capacitors can store and release electrical energy quickly, they are widely used in various practical applications.
1. Camera Flash Systems
In cameras, capacitors store electrical energy and release it in a very short time to produce a bright flash of light. The stored energy is rapidly discharged through the flash lamp to create intense illumination.
2. Power Supply Filtering
Capacitors are commonly used in power supplies to smooth voltage fluctuations. They store energy when the voltage rises and release it when the voltage drops, helping maintain a stable output voltage.
3. Energy Storage Devices
In some electronic systems, capacitors act as temporary energy storage devices. They store electrical energy and release it when needed, especially in circuits requiring rapid energy delivery.
4. Pulsed Power Systems
Capacitors are used in pulsed power systems where a large amount of energy must be released quickly. Applications include radar systems, lasers, and particle accelerators.
5. Electric Defibrillators
Medical devices such as defibrillators use capacitors to store electrical energy and release it as a controlled pulse to restore a normal heartbeat during cardiac emergencies.
6. Timing and Oscillator Circuits
In electronic circuits, capacitors store and release energy periodically, allowing them to control timing and oscillations in devices such as signal generators and clocks.
Conclusion: Energy Stored in Capacitors
In conclusion, a capacitor stores electrical energy in the electric field created between its plates when it is charged. The amount of energy stored depends on the capacitance of the capacitor and the applied voltage. This relationship is expressed by the formula $U=\frac{1}{2}CV^2$, which shows that the stored energy increases significantly as the voltage increases.
Understanding how energy is stored in capacitors is important in many areas of physics and electronics, including energy storage systems, power supplies, and electronic circuits. By studying the concepts of capacitor energy and electric field energy density, students can better understand how electrical energy is stored and transferred in electrostatic systems.
Overall, capacitors play a crucial role in modern electronic devices, making the concept of energy storage in capacitors an essential topic in electrostatics and electrical engineering.
Frequently Asked Questions (FAQ) Energy Stored in Capacitors
1. What is the energy stored in a capacitor?
Energy stored in a capacitor is the electrical energy accumulated in the electric field between the capacitor plates when the capacitor is charged. This energy results from the work done to move electric charges from one plate to the other.
2. What is the formula for energy stored in a capacitor?
The energy stored in a capacitor can be calculated using the formula:
$U=\frac{1}{2}CV^2$
where:
- U = energy stored in the capacitor (joules)
- C = capacitance (farads)
- V = potential difference across the capacitor (volts)
3. Can the energy stored in a capacitor be expressed in different forms?
Yes. The energy stored in a capacitor can be written in three equivalent forms:
$U=\frac{1}{2}CV^2$
$U=\frac{1}{2}QV$
$U=\frac{Q^2}{2V}$
where Q is the charge stored on the capacitor.
4. Where is the energy stored in a capacitor?
The energy of a charged capacitor is stored in the electric field between the plates, not directly on the plates themselves.
5. What is the energy density of an electric field?
Energy density represents the energy stored per unit volume in an electric field and is given by
$u=\frac{1}{2}\epsilon_0E^2$
where
- u = energy density (J/m³)
- $\epsilon_0$ = permittivity of free space
- E = electric field strength
6. What factors affect the energy stored in a capacitor?
The energy stored in a capacitor depends on:
- the capacitance C of the capacitor
- the voltage V applied across the plates
Since $U=\frac{1}{2}CV^2$, increasing the voltage greatly increases the stored energy.
7. What are the practical applications of capacitor energy storage?
Energy stored in capacitors is widely used in:
- power supply smoothing circuits
- camera flashes
- energy storage devices
- electronic timing circuits
- pulsed power systems
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