Content Curator
When the capacitor is being charged the electrical field tends to build up. The energy created through charging the capacitor remains in the field between the plates even after disconnecting from the charger. The amount of energy saved in a capacitor network is equal to the accumulated energies saved on a single capacitor in the network. It can be calculated as the energy saved in the equivalent capacitor of the network.
Key Terms: Capacitor, Energy, Parallel plate capacitor, Energy density, Electric charge, Electric potential, Combinations of capacitors, Potential energy, Conductors, Potentials
What is a capacitor?
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A capacitor is generally a system of two conductors that are separated from each other by an insulator.
- The conductors have charges, say Q1 and Q2, and potentials V1 and V2. Usually, if we see this practice then, the two conductors mostly have the charges as Q and – Q, with a potential difference as V = V1 – V2 between them. And to a major extent, only this kind of charge configuration shall be considered for the capacitor.
- Although, in a few cases, the single conductor could also be used significantly by assuming the other conductor at infinity.
- Hereby, Q is called the charge of the capacitor, though this could also be considered as the charge on one of the conductors as well, claiming the total charge of the capacitor as zero.
How is energy stored in a capacitor?
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Familiarity with the capacitor and its charges would help one to clearly understand the principle of energy conservation and the energy storage in a capacitor.
- Energy is stored in a capacitor because of the purpose of transferring the charges onto a conductor against the force of repulsion that is acting on the already existing charges on it.
- This work is stored in the electric field of the conductor in the form of potential energy.
- In order to determine the energy stored in the configuration of Q and -Q, we shall initially consider the two uncharged conductors A and B.
- And now, imagine the next step as a process of transferring charge from conductor B to conductor A bit–by–bit or say gradually, so that at the end, conductor A gets the charge Q. And by charge conservation, conductor 2 gets the charge –Q at the end of the entire process.
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Derivation of Energy Stored in a Capacitor
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The process of charging a capacitor is equivalent to that of transferring charges from one plate of the capacitor to another plate. Some work must be done in charging a capacitor and this work done is stored as electrostatic potential energy in the capacitor.
Let at any instant a charge q be on the plate of the capacitor. Then the potential difference between the plate is given by
V = q/C
Where C is the capacitance of the capacitor.
If extra charge dq is transferred to the capacitor, then work done to do so is stored as electric potential energy in the capacitor.
dU = dW = Vdq = q dq/C
The total increase in potential energy in charging the capacitor from q = 0 to q = Q is the total energy stored in the capacitor.
\(\Rightarrow U = \int dU = \int^Q_0 \frac {q}{C}dq\)
\(\Rightarrow U = \frac {1}{C}\int^Q_0 qdq = \frac{1}{C}[\frac {q^2}{2}]^Q_0\)
\(\Rightarrow U = \frac {1}{C}\frac {Q^2}{2}\)
Therefore, the formula of energy stored in a capacitor is given by
\(U = \frac {1}{2}\frac {Q^2}{C}\)
Substituting, Q = CV, we get
\(U = \frac {1}{2}CV^2\)
Substituting C = Q/V, we get
\(U = \frac {1}{2}QV\)
Also Read: Speed Time Graphs
Effect of Dielectric on Energy Stored in a Capacitor
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The effect of the introduction of a dielectric medium of dielectric constant K between the plates of the capacitor on energy stored in the capacitor can be calculated for two conditions. These are
- When the battery connected across the plates of the capacitor is disconnected, the energy stored in the capacitor is given by
U = Uo / K
Where Uo is the energy stored in the capacitor in the absence of a dielectric medium.
- When the battery remains connected across the plate of the capacitor, then the energy stored in the capacitor is given by
U = KUo
What is a Parallel Plate Capacitor?
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A parallel plate capacitor is said to consist of two large planes and parallel conducting plates that are simultaneously separated by a small distance. And, firstly, it is necessary to take the intervening medium between the two large planes and parallel plates to be a vacuum. Thus, representing a parallel plate capacitor.
- The direction of the electric field in this case is usually from the positive plate to the negative plate.
- Thus, the electric field is set to be localized between the two plates and is also uniform throughout.
- Taking into account the plates with the finite area, this will majorly not be considered as the option near the outer boundaries of the plates.
- The field lines, therefore, tend to bend outwards at the edges. This effect is therefore called the ‘fringing of the field’.
Distribution of Charges on the two connected Charged Capacitors:
The distribution of the charges on the connected charged capacitors differs as per the different cases of differently charged capacitors. The varying nature of the potential and the pattern of the capacitors define their distribution of charge in most cases.
Let us now see the two different cases -
(a) Common Potential:
Let us consider two plates A and C and C1 and C2 as their capacitors along with Q1 and Q2 as their charges respectively.
Conservation of charge is provided on plates A and C before and after connection.
Q1 + Q2 = C1 V + C2 V
Therefore, V = Q1 + Q2 / C1 + C2 = C1V1 + C1V2 / C1 + C2
Or,
It could also be said as in other words the common potential will be equal to the total charge on the capacitors divided by the total capacitance of the system.
(b) We use the following method in order to obtain the values of the final charge on either of the capacitors:
Here, we can see that the ‘V’ is the common potential. Similarly, it could also be written as :
Therefore, the heat lost during the redistribution of the charges is :
Important points to be noted:-
- When the plates of similar charges are connected with each other ( + with + and – with -) then all the values are required to be fetched (Q1, Q2, V1, V2) with a positive sign.
- When the plates of the opposite polarity are connected with each other ( + with -) then it is essential to take the charge and potential of one of the plates to be negative to get the correct and apt. outcome.
Electrostatic Energy Density of a Parallel Plate Capacitor
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Consider a parallel plate capacitor of plate area A and the distance of separation between the plates d. The energy stored in the capacitor is given by
U = CV2/2 …(i)
The capacitance of a parallel plate capacitor is given by
C = εoA/d
Also, we have V = Ed
Substituting the above values in equation (i), we get
\(U=\frac {1}{2} \frac {\epsilon_oA}{d} \times E^2d^2\)
\(\Rightarrow U=\frac {1}{2} \epsilon_oE^2Ad\)
But the volume of the capacitor is Ad
\(\Rightarrow \frac {U}{Ad}=\frac {1}{2} \epsilon_oE^2\)
Energy stored per unit volume of a capacitor is known as Energy density.
Therefore, the Energy Density of a capacitor is given by
\(U_d=\frac {1}{2} \epsilon_oE^2\)
The unit of energy density is Jm-3 and the dimensional formula is [ML-1T-2].
Energy Stored in a Combination of Capacitors
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Two or more capacitors can be combined in two ways
- Series combinations of capacitors
- Parallel combination of capacitors
Let U1, U2, U3,……...Un be the energy stored in n number of capacitors. If they are connected in series or in parallel combinations, then the energy stored in the combinations of capacitors is given by
U = U1 + U2 + U3 +……..+ Un
Also Read: Angular Speed
Loss of Energy on Sharing Charges
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When two or more capacitors are connected in parallel, no charges are lost but some energy is lost on sharing the charges between two capacitors. This loss of energy appears as heat energy.
Consider two capacitors of capacitance C1 and C2 are at potential V1 and V2 respectively. When they are connected in parallel, then the loss of energy that appears as heat is given by
\(H = \frac {C_1C_2 (V_1-V_2)^2}{2(C_1+C_2)}\)
Solved Examples
Que. If the capacitance of a capacitor is 50 F charged to a potential of 100 V, Calculate the energy stored in it.
Sol. We have a capacitor that has a capacitance of 50 F and is charged to a potential of 100 V. The energy which is stored in the capacitor can be calculated in the following manner –
U = 1/2 CV2
Now, while substituting the values, we get
U = 1/2 x 50 (100)2 = 250 × 10 3 J
Que. The potential difference across the plates of a parallel plate capacitor is 50 V and the charge of the plate is 2 C. Find the energy stored in the capacitor.
Sol. Given
- The potential across the plates, V = 50 V
- Charge, Q = 2C
Energy stored in a capacitor is given by
U = 1/2 QV
Now, while substituting the values, we get
U = 1/2 x 2 x 50 = 50 J
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Things to Remember
- A capacitor is generally a system of two conductors that are separated from each other by an insulator.
- Familiarity with the capacitor and its charges would help one to clearly understand the principle of energy conservation and the energy storage in a capacitor.
- Energy is stored in a capacitor because of the purpose of transferring the charges onto a conductor against the force of repulsion that is acting on the already existing charges on it.
- This work is stored in the electric field of the conductor in the form of potential energy.
- A parallel plate capacitor is said to consist of two large planes and parallel conducting plates that are simultaneously separated by a small distance.
- The direction of the electric field in this case is usually from the positive plate to the negative plate.
- The distribution of the charges on the connected charged capacitors differs as per the different cases of differently charged capacitors.
Sample Questions
Ques. What can be the possible applications for energy capacitors? (3 Marks)
Ans. Following are a few applications of capacitor energy:
- A defibrillator that is used to correct abnormal heart rhythm delivers a large charge in a short burst to a person’s heart. When large shocks of electric current are applied it helps in the stopping of arrhythmia and allows the natural pacemaker of the body to regain its usual rhythm. The energy stored in the capacitor is made use of by the defibrillator.
- There are various other pieces of equipment that use energy stored in a capacitor like audio equipment, uninterruptible power supplies, camera flashes, and pulsed loads such as magnetic coils and lasers.
Ques. What are conductors and insulators? (1 Mark)
Ans. Conductors contain a large number of free-charge carriers to conduct electricity while insulators do not contain any free-charge carriers to conduct electricity.
Ques. In what fields is this topic useful? (1 Mark)
Ans. This topic is necessary for individuals pursuing their careers in physics and engineering at a higher level.
Ques. (a) Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.
(b) A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor. (5 Marks)
Ans.
(a) Let us think that between the plates we have a capacitor of capacitance and a potential difference V and the charge is +q on one and -q on another. Let’s consider that the capacitor is charging moderately and then at any stage q is the charge on the capacitor. Then the possible distinction can be = q/C. The little amount of work done by the additional charge dq to the capacitor refers to:
dW = (q/C)*dq
The complete amount of work done in providing a charge Q to the capacitor refers to W = ∫dW
W = 0∫Q (Q/C)dq
W = Q²/C
Energy = E
E = Q²/2C = CV²/2 = QV/2
The energy can be saved in the form of potential energy.
(b) The initial saved energy, Ui = ½ CV²
The charge can be seen to distribute equally when the uncharged capacitor is connected. Since, Q = CV, Vf = V/2
Therefore energy can be stored in two capacitors that can be shown:
Uf = 2*½ CVf²
Uf = CV²/4
Therefore the ratio that states the energy saved in the combined system to the one saved in the beginning: Uf/Ui = ½
Ques. A capacitor of capacitance C1 is charged to a potential V1 while another capacitor of capacitance C2 is charged to a potential difference V2. The capacitors are now disconnected from their respective charging batteries and connected in parallel to each other.
(a) Find the total energy stored in the two capacitors before they are connected.
(b) Find the total energy stored in the parallel combination of the two capacitors.
(c) Explain the reason for the difference of energy in parallel combination in comparison to the total energy before they are connected. (5 Marks)
Ans.
(a) The initial total energy:
C = ½ C1 V1 ² + ½ C2 V2 ²
(b) If v is referred to as the potential across the parallel compound. Then when we apply the conservation of charge, we can say:
(C1 +C2 )V = C1 V1 + C2 V2
V = (C1 V1 +C2 V2 ) / C1 +C2
The complete energy saved in the parallel compound:
E = ½ (C1 +C2 ) [(C1 V1 +C2 V2 ) / C1 +C2 ]²
= ½ [(C1 V1 +C2 V2 )² / C1 +C2 ]
(c) We can see the difference in energy because of the loss of energy that can be due to the motion of charge when the charge is being shared.
Ques. Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric. (5 Marks)
Ans. The initial charge on each capacitor = QA = QB = CV
After the introduction of the dielectric, the new capacitance of each capacitor = KC
Changes to V volts after the opening of the switching potential across the capacitor.
If we consider the potential across the capacitor B as V1
Then QB = CV = C1 V1 = KCV1
V1 = V/K
The initial energy in the two capacitors = CV²/2 + CV²/2 = CV²
Then the resultant energy of capacitor A = KCV²/2
Then the resultant energy of capacitor B = KCV²/2K² = CV²/2K
The total amount of energy in the end of both the capacitors = KCV²/2 + CV²/2K (K²+1/2K)CV²
The ratio of the total amount of the electrostatic energy saved in the two capacitors earlier and later the introduction of the dielectric = CV² / [(K²+1) / 2K]CV² = 2K / (K²+1)
Ques. (a) Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.
(b) Two charged spherical conductors of radii R1 and Ri when connected by a conducting wire acquire charges q1 and q2 respectively. Find the ratio of their surface charge densities in terms of their radii. (5 Marks)
Ans.
(a) The electric field present inside the parallel plate capacitor because one of the plates is represented by:
E1 = σ/2ε0 = Q/2Aε0
The complete electric field has given because of two plates is represented by:
E = 2E1 = Q/Aε0
The possible difference between the two plates is:
V = Ed
V = Qd/Aε0
By explanation, C = Q/V =Aε0/d
(b) After connecting, two of the spheres are at a similar possibility:
V1 = V2
Kq1 /R1 = Kq2 / R2
q1 / q2 = R1 / R2
The density of the surface charge for a circular conductor is:
σ2= Q/S
σ1 = q1/4πR1²
σ2 = q2/4πR2²
σ1 / σ2 = q1R12 = R1 / R2
Ques. Two parallel plate capacitors X and Y have the same area of plates and the same separation between them. X has air between the plates while Y contains a dielectric medium of ε r = 4.
(i) Calculate the capacitance of each capacitor if the equivalent capacity of the combination is 4 µF.
(ii) Calculate the potential difference between the plates of X and Y.
(iii) Estimate the ratio of electrostatic energy stored in X and Y. (5 Marks)
Ans.
(i) Let's consider Cx = C
Cy = 4C (dielectric medium of r = 4)
For the combination series of the two capacitors:
1/C = 1/Cx + 1/Cy
1/4μF = 1/C + 1/4C
1/4μF = 5/4C
C = 5μF
Therefore Cx = 5μF
Cy = 20μF
(ii) Total charge Q = CV
= 4μF * 15V = 60μC
Vx = Q/Cx = 60μF/5μF = 12V
Vy = Q/Cy = 6μF/20μF = 3V
(iii) Ex/Ey = (Q²/2Cx) / (Q²/2Cy)
= Cy/Cx = 20/5
= 4:1
Ques. (a) Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor.
(b) The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving a charge q over a closed rectangular loop a b c d a (3 Marks)
Ans.
(a) The energy saved in a capacitor can be demonstrated by:
E = ½ CV²
When we use C = (ε0 A) / d
E = ½ ε0 Ad(V²/d²)
When we use the possible gradient, E = V/d
E/Ad = ½ ε0 E²
The term used on the left-hand side is referred to as the energy that is saved per unit volume in the capacitor.
(b) The network done by the electric force present in the motion that helps in moving a charge through a loop is zero due to the electric field being unchanging.
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