The use of capacitors is very common in various devices like handheld electronic calculators, railway fans, etc. But how is energy stored in a capacitor? In this context, you will get to know how a capacitor holds energy, along with the calculation of the same.
What is a Capacitor?
The capacitor is an electrical energy storing device. Additionally, most capacitors contain two terminals located side by side while an insulator is present between them. In some cases, this whole unit is modified into a compact device in order to save space. Moreover, there are few capacitors which have multiple layers in them for additional functionalities.
How Does a Capacitor Hold Energy?
Two positive charges cannot do anything with one another. Instead, they move away from each other as quickly as they can. However, if the two charges are made to come closer forcibly, they resist. Also, it requires energy to make them come close.
Furthermore, the energy needed does not stray or get utilised. Rather, it gets stored in the form of an electric field which is a type of tension; provided the charges are clasped together, uncomfortably.
Moreover, when the charges again have the liberty to move, they utilise energy to speed them up. Thus, it can be said that capacitors are those components that store electric fields.
Evaluation of Energy Stored in a Capacitor
Let us consider a capacitor is charged to a certain amount of voltage V, and its energy is needed to be calculated. So, energy (or work) W required to move a positive charge close to another one is the product of the positive charge Q and voltage (potential difference).
δW = Q x δV
(joules)=(coulombs)x(volts)
However, as per common logic, some individuals may feel that a capacitor with charge V needs energy of QV joules to reach the desired state, and hence the capacitor is holding QV joules of energy. However, that is not the case.
Instead, as the charges move nearer and nearer to each other, their resisting property keeps on increasing till it becomes fierce. It is a non-linear procedure. Hence, the only process for energy stored in a capacitor derivation is using the method of integration.
For example, assume that capacitor C is storing a charge Q. So, measuring the voltage V across it can be done quite easily. Further, after applying a small amount of energy, a bit of charge can be induced to the system. Therefore, in terms of Q, an expression can be written.
δW = V δQ = \[\frac{Q}{C}\] δQ
After understanding this equation, by integration of the complete δW, requiring energy to push charge Q to the capacitor can also be evaluated.
W=\[\frac{1}{C}\]\[\int_{0}^{Q}\]QdQ = \[\frac{1}{C}\] \[\frac{Q²}{2}\]= \[\frac{1}{2}QV\]
Take a look at the below expression for energy stored in capacitor.
W = \[\frac{1}{2}\]CV² (joules)
Moreover, here is a solved numerical which will make you understand the calculation better.
Numerical
(i) A capacitor has a capacitance of 50F and it has a charge of 100V. Find the energy that this capacitor holds.
Solution. According to the capacitor energy formula:
U = 1/ 2 (CV2)
So, after putting the values:
U = ½ x 50 x (100)2 = 250 x 103 J
Do It Yourself
1. The Amount of Work Done in a Capacitor which is in a Charging State is:
(a) QV (b) ½ QV (c) 2QV (d) QV2
By going through this content, you must have understood how capacitor stores energy. Additionally, for more knowledge about capacitors, circuits, and other concepts of Physics, download our Vedantu app. Along with easy access to study materials; it also offers online interactive sessions for better understanding of these topics.
FAQs on Energy Stored in a Capacitor
1. What is meant by the energy stored in a capacitor?
The energy stored in a capacitor is the electric potential energy gained during the charging process. This energy comes from the work done by an external source, like a battery, to move charge from one plate to another against the electrostatic forces. This stored energy is then held within the electric field between the plates and can be released when the capacitor is discharged.
2. What are the main formulas for calculating the energy stored in a capacitor?
The energy (U) stored in a capacitor can be calculated using three primary formulas, based on the known variables of charge (Q), potential difference (V), and capacitance (C):
- U = ½ (Q²/C): Use this when you know the charge and capacitance.
- U = ½ CV²: This is the most common formula, used when capacitance and voltage are known.
- U = ½ QV: Use this when you know the charge and the potential difference.
All three formulas are equivalent and can be derived from one another using the core relationship Q = CV.
3. How is the formula for the energy stored in a capacitor, U = ½ CV², derived?
The formula is derived by calculating the total work done (W) to charge a capacitor. We can think of this process in small steps:
- At any point during charging, let the instantaneous charge be 'q' and the potential difference be V = q/C.
- The small amount of work (dW) needed to add another small charge 'dq' is given by dW = V dq = (q/C) dq.
- To find the total work to charge the capacitor from zero to a final charge Q, we integrate this expression: W = ∫(from 0 to Q) (q/C) dq.
- Solving the integral gives W = [q²/2C] from 0 to Q, which results in W = Q²/2C.
Since this work done is stored as potential energy (U), we have U = Q²/2C. By substituting Q = CV, we get the popular form U = ½ CV².
4. In what form is energy stored in a capacitor, and where exactly is it located?
A charged capacitor stores energy in the form of electric potential energy. Contrary to what some may think, this energy is not stored on the metal plates themselves. Instead, it is stored in the electric field that is established in the volume of the dielectric material (like air, paper, or ceramic) that separates the two plates.
5. How does the energy stored in a capacitor change if a dielectric slab is introduced between its plates while it is still connected to the battery?
If the battery remains connected, the potential difference (V) across the capacitor stays constant. Here's the impact:
- Introducing a dielectric slab with dielectric constant 'K' increases the capacitance from C to C' = KC.
- The energy stored is given by the formula U = ½ CV².
- The new energy will be U' = ½ C'V² = ½ (KC)V² = K(½ CV²).
- Therefore, the new energy is U' = K × U.
Since K > 1 for all dielectrics, the energy stored in the capacitor increases by a factor of K. The extra energy is supplied by the battery to move more charge onto the plates.
6. How is the energy stored in a capacitor related to the electric field between its plates?
The energy in a capacitor is fundamentally stored in its electric field. This relationship is defined by the concept of energy density (u), which is the energy stored per unit volume. The formula for energy density is:
u = ½ ε₀E²
Here, 'E' is the magnitude of the electric field and 'ε₀' is the permittivity of free space. The total energy (U) is this energy density multiplied by the volume (Area × distance) of the space between the plates, proving that the energy resides entirely within the field.
7. Why does the energy stored in an isolated charged capacitor decrease when a dielectric is inserted?
When a capacitor is isolated, its charge (Q) remains constant because it's not connected to a battery. The initial energy is U = Q²/2C. When a dielectric (constant K) is inserted, the capacitance increases to C' = KC. The new energy becomes U' = Q²/2C' = Q²/(2KC) = (1/K) * (Q²/2C) = U/K. Since K > 1, the final energy U' is less than the initial energy U. This energy reduction occurs because the electric field inside the capacitor does work on the dielectric, pulling the slab into the space between the plates.
8. How do you calculate the total energy stored in a combination of capacitors in series versus in parallel?
The total energy is always the sum of the energies in individual capacitors, but the calculation method differs based on the connection type:
- For capacitors in parallel: The voltage (V) across each is the same. The total energy is the sum of individual energies: U_total = ½ C₁V² + ½ C₂V² + ... = ½ (C₁ + C₂ + ...)V². This is the energy of the equivalent parallel capacitance.
- For capacitors in series: The charge (Q) on each is the same. The total energy is again the sum: U_total = Q²/2C₁ + Q²/2C₂ + ... = (Q²/2)(1/C₁ + 1/C₂ + ...). This is the energy of the equivalent series capacitance.
