Capacitors are commonly used to store electrical energy and release it when needed. They store energy in the form of electrical potential energy.
How do capacitors store energy?
Capacitance is the ability of a capacitor to store charge, which is measured in Farad. Capacitors are usually used in conjunction with other circuit components to produce a filter that allows some electrical impulses to pass while blocking others.
Figure 1. Capacitors
Capacitors are made of two conductive plates and an insulator material in between them. When a capacitor is connected to a circuit, the positive pole of the voltage source begins to push the electrons from the plate to which it is connected. These pushed electrons gather in the other plate of the capacitor, causing excess electrons to be stored in the plate.
Figure 2. Diagram of a charged capacitor. Source: Oğulcan Tezcan, StudySmarter.
The excess electrons in one plate and their corresponding lack in the other cause a potential energy difference (voltage difference) between the plates. Ideally, this potential energy difference (charge) remains unless the capacitor begins to discharge in order to supply voltage back to the circuit.
However, in practice, there are no ideal conditions, and the capacitor will begin to lose its energy once it is taken out of the circuit. This is because of what is known as leakage currents out of the capacitor, which is an unwanted discharging of the capacitor.
The effect of the dielectric on the stored charge
How long a capacitor can store energy depends on the quality of the dielectric material between the plates. This insulating material is also known as the dielectric. How much energy a capacitor stores (its capacitance) is decided by the surface area of the conductive plates, the distance between them, and the dielectric between them, which is expressed as follows:
\[C = \frac{\epsilon_0 \cdot A}{d}\]
Here:
- C is capacitance, measured in Farad.
- \(\epsilon_0\) is the dielectric constant of the insulator material.
- A is the area of plate overlap (\(m ^ 2\)).
- d is the distance between the plates, measured in metres.
The table below indicates how much effect the dielectric material has on the energy stored by the capacitor.
| Material | Dielectric constant |
| Air | 1.0 |
| Glass (window) | 7.6-8 |
| Fibre | 5-7.5 |
| Polyethylene | 2.3 |
| Bakelite | 4.4-5.4 |
How to calculate the energy stored in a capacitor
Since the energy stored in a capacitor is electrical potential energy, it is related to the charge (Q) and the voltage (V) of the capacitor. First, let’s remember the equation for electrical potential energy (ΔPE), which is:
\[\Delta PE = q \cdot \Delta V\]
This equation is used for the potential energy (ΔPE) of a charge (q) while going through a voltage difference (ΔV). When the first charge is placed in the capacitor, it goes through a change of ΔV=0 because the capacitor has zero voltage when it is not charged.
When the capacitor is fully charged, the final charge being stored in the capacitor experiences a voltage change of ΔV=V. The average voltage on a capacitor during the charging process is V/2, which is also the average voltage experienced by the final charge.
\[E_{cap} = \frac{Q \cdot V}{2}\]
Here:
- \(E_{cap}\) is the energy stored in a capacitor, measured in Joules.
- Q is the charge on a capacitor, measured in Coulombs.
- V is the voltage on the capacitor, measured in Volts.
We can express this equation in different ways. The charge on a capacitor is found from the equation Q = C*V, where C is the capacitance of the capacitor in Farads. If we put this into the last equation, we get:
\[E_{cap} = \frac{Q \cdot V}{2} = \frac{C \cdot V^2}{2} = \frac{Q^2}{2 \cdot C}\]
Now, let’s consider some examples.
A heart defibrillator is giving out \(6.00 \cdot 10^2\) J of energy by discharging a capacitor, which initially is at \(1.00 \cdot 10 ^ 3\) V. Determine the capacitance of the capacitor.
The energy of the capacitor (Ecap) and its voltage (V) are known. As we need to determine the capacitance, we need to use the relevant equation:
\[E_{cap} = \frac{C \cdot V^2}{2}\]
Solving for the capacitance (C), we get:
\[C = \frac{2 \cdot E_{cap}}{V^2}\]
Adding the known variables, we then have:
\[C = \frac{2 \cdot (6.00 \cdot 10^2 [J])}{(1.00 \cdot 10^3 [V])^2} = 1.2 \cdot 10^{-3} [F]\]
\(C = 1.2 [mF]\)
The capacitance of a capacitor is known to be 2.5 mF, while its charge is 5 Coulombs. Determine the energy stored in the capacitor.
As the charge (Q) and the capacitance (C) are given, we apply the following equation:
\[E_{cap} = \frac{Q^2}{2 \cdot C}\]
Adding the known variables, we get:
\[E_{cap} = \frac{(5[C])^2}{2 \cdot (2.5 \cdot 10^{-3} [F])}= 5000 [J]\]
\(E_{cap} = 5 [kJ]\)
Energy Stored by a Capacitor - Key takeaways
- Capacitance is the storing ability of a capacitor, which is measured in Farad.
- How long a capacitor can store energy is determined by the quality of the insulator material (dielectric) between the plates.
- How much energy a capacitor stores (its capacitance) is determined by the surface area of the conductive plates, the distance between them, and the dielectric between them.
- The equation used to determine the capacitance is \(C = \frac{(\epsilon_0 \cdot A)}{d}\).
- The equation used to determine the energy stored in the capacitor is \(E = \frac{Q \cdot V}{2}\).
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