Parallel Plate Capacitor: Derivation, Construction, Working, Application & Physics Notes

A parallel plate capacitor stores both electric charge and electric energy with two flat conducting plates that lie parallel to each other. The gap between the plates has an insulation material known as the dielectric.
The parallel plate capacitor works when we apply a voltage to the plates. That causes them to become positively and negatively charged, which leads to creating an electric field. This field is uniform and is able to store potential energy to be further used for release when the circuit needs it.
In physics, we call this ability to store a charge as the capacitance.
To define in mathematical terms, we can use the following parallel plate capacitor formula.

C = εA / d

Here,
C remains as the capacitance
ε tells how much the dielectric material permits
A is the area of the material

There are two conductor plates, one dielectric material, two supports for the insulation to happen, and two terminals.

1. Two Conductor Plates

These two are of copper in this setup. Otherwise, brass or aluminium would work. These plates have to be thin and flat, and both must lie parallel to each other.
One thing to note here is that a larger surface would be able to store more charge.

2. Dielectric Material

The insulating material between the two conducting places is a paper, glass, or mica. This material is placed here so that there is no direct contact.

Increases the capacitance by reducing the electric field leakage.

3. Support and Insulation

To hold the plates in place, insulating supports come in to maintain the fixed gap.
In this setup, the distance between the plates is kept small for high capacitance but large to avoid breakdown.

4. Terminals

Each plate is attached with metallic connectors.
The terminal is connected to an external circuit or power source.

Below is the step-by-step working of a parallel plate capacitor. 

When a voltage is applied to the capacitor, electrons are pulled from the positive plate and pushed to the negative plate. The plates become positively and negatively charged. The dielectric materials prevent the flow of charges between the plates. A uniform electric field exists between the plates due to the opposite charge. This electric field stores the potential energy.

Also Read: NCERT Solutions for Class 11 & 12

The energy is stored in the capacitor until it is connected to a circuit.  The charge stored in the capacitor flows from the negative to the positive plate when it is attached to a load. 

The capability of a capacitor to store charge is its capacitance (C). This is based on the distance (d) between the charge, plate area (A) and the permittivity of the dielectric material.

The formula of a parallel plate capacitor is expressed as:  

Capacitance (C) = εA / d 

Here, 

• C is the capacitance   
• ε is the permittivity of the dielectric material 
• A is the area of a plate 
• d is the distance between the plates

Important Links:

NCERT Class 11 notes	CBSE Class 11 Physics notes
Class 11 Chemistry notes	NCERT Class 11 Maths notes

When a plate is connected to a voltage source (battery), the plate gets a positive charge (+Q) and the other gets a negative charge (- Q). The electric field for an infinite charged sheet is:

E = σ / 2ε 

Here, σ = Q / A (surface density) 

The electric field between the plates

E =  σ / ε = Q / εA

Potential Difference between the plates

The formula for potential difference is: V = E * D 

 V = Q / εA * d 

Capacitance Formula

 C = Q / V 

C = Q / Q / εA * d 

Simply C = εA / d 

Including the dielectric constant

If a dielectric material has relative permittivity εr, then  

 ε = ε0εr 

C = ε0εrA / d 

The final formula for the capacitance of a parallel plate capacitor is given as: C = εA / d or C = ε0εrA/d

We find uses for parallel plate capacitors in several real-life scenarios beyond the laboratories.

1. For Flash Photography

Parallel plate capacitors use the stored energy to quickly discharge as and when some equipment needs it. These capacitors have a high capacitance.
The best use case is for camera flashes, where the battery stores the energy in the form of electrical charge. When you click the button, large amount of energy releases in a matter of seconds.

2. In Medical Science

Parallel plate capacitors have both quick and rapid charge and discharge capabilities. For that, we see its mechanism work in a defibrillator. This medical device is used to restore a heartbeat by providing an electric shock and helps treat arrhythmias. It has a defined quantity of electrical charge.

 

Example 1: A parallel plate capacitor has plates of area $0.02{\text{m}}^2$ separated by 1 mm. Calculate the capacitance with air (𝜀𝑟=1) and with a dielectric (𝜀𝑟=3).

Solution:

For air: 

$C=\frac{{\epsilon }_{0}A}{d}=\frac{(8.85\times {10}^{-12})·0.02}{0.001}=1.77\times {10}^{-10}\text{F}=177\text{pF}$

With dielectric: 

$C=\frac{{\epsilon }_r{\epsilon }_{0}A}{d}=3·1.77\times {10}^{-10}=5.31\times {10}^{-10}\text{F}=531\text{pF}$

Example 2: A parallel plate capacitor with 𝐶=100pF is charged to 200 V. A dielectric slab ( 𝜀𝑟=2 ) is inserted with the battery disconnected. Calculate the new potential difference and energy stored.

Solution: 

Initial charge: $Q=CV=(100\times {10}^{-12})·200=2\times {10}^{-8}\text{C}$

New capacitance: $C^{\text{'}}={\epsilon }_{r}C=2·100\times {10}^{-12}=200\text{pF}$

New potential difference: $V^{\text{'}}=\frac{Q}{C^{\text{'}}}=\frac{2\times {10}^{-8}}{200\times {10}^{-12}}=100\text{V}$

Initial energy: $U=\frac{1}{2}C{V^{\text{'}}}^2=\frac{1}{2}(100\times {10}^{-12})·{\left(200\right)}^2=2\times {10}^{-6}\text{J}$

New energy: $U^{\text{'}}=\frac{1}{2}\frac{Q^2}{C^{\text{'}}}=\frac{1}{2}\frac{{(2\times {10}^{-8})}^2}{200\times {10}^{-12}}=1\times {10}^{-6}\text{J}$

The energy decreases due to work done against polarization forces.