# Derive and Expression for the Capacitance of a Parallel Plate Capacitor with and without Dielectric Medium between the Plates

Parallel plate capacitors have already been addressed. But do you have any idea what they are? Is it a configuration in which two plates are parallel to one another? Why don’t you try to figure it out for yourself? To understand the concept of the parallel plate capacitor, read this article.

### The Parallel Plate Capacitor

Parallel Plate Capacitors are capacitors with electrodes and insulating material arranged in a parallel pattern (dielectric). The electrodes are the two conducting plates. Between them, there is a dielectric. For the plates, this serves as a divider.

The parallel plate capacitor’s two plates are of similar size. They are plugged into the power source. The plate, which is linked to the battery’s positive terminal, receives a positive charge. The plate linked to the negative terminal of the battery, on the other hand, receives a negative charge. Charges are trapped inside the capacitor plates as a result of the attraction.

### The Principle of Parallel Plate Capacitor

We know that we can charge a plate to a particular level. If we provide additional charge, the potential grows, which might result in a charge leakage. When we set another plate next to this positively charged plate, negative charge flows to the side of the plate that is closest to the positively charged plate.

The negative charge on plate 2 will lower the potential difference on plate 1 since both plates have charges. The positive charge on plate 2 on the other hand, will increase the potential difference on plate 1. Plate 2’s negative charge, on the other hand, will have a greater influence. As a result, plate 1 can get a higher charge. Because plate 2 has negative charges, the potential difference will be less. The parallel plate capacitor works on this basis.

### Dependence of Charge Stored in a Capacitor

The potential difference between the two plates of a parallel plate capacitor is exactly proportional to the amount of electric charge stored in any of the plates. This relationship may be deduced as follows:

**Q ∝ V**

Therefore, **Q = (constant) V = C V**

Where

- V is the potential difference between the plates,
- Q is the amount of charge stored, and
- C is the capacitance of the capacitor.

**The Capacitance of Parallel Plate Capacitor**

The quantity of charge that a parallel plate capacitor can retain is determined by its capacitance. If you look at the following equation, you can see that the higher the value of C, the more charge a capacitor can retain. As a result, we can see that capacitance is determined by:

- The area A of the medium between plates &
- The distance d between plates.

According to Gauss Law, the electric field is given as:

E = Q ⁄ ε_{0}AThe potential difference between plates is given as:

V = E d = Q d ⁄ ε_{0}ASince, the capacitance is defined as C = Q ⁄ V, so formula of capacitance can be given as:

C = ε_{0}A ⁄ dThe greatest capacitance is obtained when the plates are positioned extremely close together and the area of the plates is big.

### Dielectric Material Inserted between Two Plates

The charges will be shielded on the two plates by the material’s tiny dipole moment. As a result, the impact of the dielectric substance put between the two plates will be altered. The relative permeability k determines the permeability of a material. The capacitance is calculated as follows:

**C = ε A ⁄ d = k ε _{0} A ⁄ d**

The capacitance of a parallel plate capacitor can be increased by adding a dielectric between the plates with a permeability k higher than 1. Dielectric Constant is another name for K.

### Multiple Parallel Plate Capacitor

Multiple Parallel Plate Capacitor is an arrangement of parallel plate capacitors with dielectric material between them in groups that fit together. The capacitance of a capacitor with numerous parallel plates may be computed as follows:

C = [ε_{0}ε_{r}A ⁄ d](N − 1)Where

- N is the number of plates,
- d is the distance between plates,
- ε
_{r}is the relative permittivity of dielectric,- ε
_{0}is the relative permittivity of a vacuum, and- A is the area of each plate.

### Sample Problems

**Problem 1: A parallel plate capacitor is kept in the air has an area of 0.25 m ^{2} and separated from each other by distance 0.08m. Calculate the capacitance of parallel plate capacitor.**

**Solution:**

Given:

Area of plates, A = 0.25 m

^{2}Distance between plates, d = 0.08 m

Dielectric constant, k = 1

Relative permittivity, ε

_{0}= 8.854 × 10^{−12}F ⁄ mThe parallel plate capacitor formula is expressed by,

C = k ε

_{0}A ⁄ d= 1 × 8.854 × 10

^{−12}× 0.25 / 0.08= 27.67 × 10

^{−12}FHence, the capacitance of the given capacitor is

27.67 × 10.^{−12}F

**Problem 2: Determine the area of parallel plate capacitor in the air if the capacitance is 15 nF and separation between the plates is 0.02m.**

**Solution:**

Given:

Capacitance, C = 15 nF

Distance between plates, d = 0.02 m

Dielectric constant, k = 1

Relative permittivity, ε

_{0}= 8.854 × 10^{−12}F ⁄ mThe parallel plate capacitor formula is expressed by,

C = k ε

_{0}A ⁄ dA = C d ⁄ k ε

_{0}= 0.02 × 15 ×10

^{−9}/ 1 × 8.854 × 10^{−12}≈ 34 m

^{2}Hence, area of parallel plate capacitor is

34 m.^{2}

**Problem 3: Derive the expression for capacitance of parallel plate capacitor.**

**Solution:**

Since, the capacitor is made up of parallel thin plates:

Electric field by a single thin plate, E′ = σ ⁄ 2ε

_{0}Total electric field between the plates, E = σ ⁄ 2ε

_{0}+ σ ⁄ 2ε_{0}E = σ ⁄ ε

_{0}E = Q ⁄ ε

_{0}AThe potential difference between plates is given as:

V = E d = Q d ⁄ ε

_{0}ACapacitance of a capacitor is given as:

C = Q ⁄ V

Hence, the capacitance of parallel plate capacitor is

C = ε._{0}A ⁄ d

**Problem 4: A capacitor has a capacitance of C when a potential difference of V is across it. The potential difference is increased to 3V, what is the new capacitance?**

**Solution:**

The capacitance of a capacitor is given by:

C = Q ⁄ V, where

- Q is the charge on capacitor, and
- V is the potential difference across the plates.
Charge on capacitors rises as the potential difference V increases, so the ratio Q/V remains constant, i.e. capacitance remains constant. Because capacitance is dependent on plate area, medium between plates, and distance between plates, capacitance will be C when the potential difference is increased to 3V.

Hence, the new capacitance of capacitor is

C.

**Problem 5: How is the electric field between the two plates of a parallel plate capacitor?**

**Solution:**

The electric field is directed from the positive to the negative plate. The electric fields generated by the two charged plates build up in the inner area, between the two capacitor plates. As a result, the playing field is consistent throughout.

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