A simple note on the Parallel Plate Capacitor

A device or a passive electronic component with two terminals that stores electrical energy in an electric field is called the capacitor. The effect marked by the capacitor is the capacitance of the parallel plate capacitor. As discussed above, a parallel plate capacitor comprises two metal plates arranged in a parallel manner at some distance from one another. This distance comprises any dielectric medium (an insulating medium that cannot conduct electric current). Some examples of this dielectric medium include air, vacuum, mica, paper wool, electrolytic gel, glass etc. The dielectric medium is non-conducting, however, under the effect of the electric field, they do get polarised. This causes the production of dipoles and therefore both charges i.e. a negative and a positive charge gets deposited over the plates of the capacitor. Let’s study in brief to know about the formula of parallel plate capacitors and the capacitance of parallel plate capacitors. 

Parallel Plate Capacitor

As already discussed, the parallel plate capacitors comprise two conducting plates, these conducting plates act as electrodes. The dielectric medium present between these plates acts as a separator. The plates are of equal dimensions connected to the power supply. They acquire charges based on connection i.e. a positive terminal of the battery acquires a positive charge whereas the negative terminal of the battery acquires a negative charge. A specific amount of charge is to be supplied; in case of a greater charge the potential increases and causes leakage in the charge. If we place another positively charged plate near this plate, then the negative charge starts flowing towards the nearer positive charged plate. The negative charge reduces the potential difference of plate 1 i.e. the positively charged plate. However, the positively charged plate increases the potential difference between the negatively charged plate. The effect of the negative charge plate is greater than the positive charge plate, therefore, there will be a lesser potential difference. 

The Formula for Parallel Plate Capacitor

Suppose a parallel-plate capacitor comprising two metallic plates has an area of A. Moreover, the distance separating these two plates is d. The formula for a parallel plate capacitor will be as follows:

C=kϵoA/d

Here, ϵo =permittivity of space, whose value is 8.854 × 10−12 F/m

k =relative permittivity of dielectric material

d= distance of separation between the plates

A= area of plates

Factors on Which the Charge Stored in a Capacitor Depend

The potential difference between two plates is directly proportional to the amount of electric charge stored in the plates of the capacitor (parallel plate capacitor). This simply means that Q ∝ V. This can also be written as Q= C V.

C= capacitance

Q= charge

And V=Potential difference

According to the parallel plate capacitor formula with dielectric, C=kϵoA/d, the value of C is directly proportional to A (area) whereas it is inversely proportional to d (distance between the plates).

Solved Questions on Parallel Plate Capacitor 

Q 1- Assume a parallel plate capacitor has plates of side 5 cm. These plates are separated by a distance of 1 mm. Determine the capacitance of this capacitor. Moreover, calculate the charge stored in any of these plates if a 10 V battery is connected to the capacitor. (εo = 8.85 x 10-12 Nm2 C-2)

Solution:

By using the formula for parallel plate capacitor, 

C=kϵoA/d

Substituting the values, we will get

C = 221.2 ×10−13 F

C = 22 . 12 ×10−12 F

C = 22 .12 pF

The charge stored in any of these plates if a 10 V battery is connected to the capacitor will be as follows:

Q = CV

= 22. 12 ×10−12 ×10 

= 221.2 ×10−12 C 

= 221.2 pC

Q 2- Assume a parallel plate capacitor with an area of 0.50 m2 placed in air. The distance between the plates is 0.04m. Use the formula of a parallel plate capacitor to determine its capacitance. 

Solution:

According to the question,

A = 0.50 m2,

d = 0.04 m,

k = 1

ϵo = 8.854 × 10−12 F/m

By using the formula for parallel plate capacitor, 

C= 8.854×10−12 × 0.50 / 0.04

C = 4.427 x 10−12 / 0.04

C = 110.67 x 10−12 F

Q 3- The capacitance of a parallel plate capacitor is given as 25 nF and the distance between the plates is 0.04m. Determine the area of this parallel plate capacitor by using the formula. 

Solution:

According to the question:

C= 25 nF,

d = 0.04 m,

k = 1,

ϵo = 8.854 × 10−12 F/m

By using the formula for parallel plate capacitor, 

C=kϵoA/d 

Re arranging the formula for area and substituting the value will give us,

Area= 0.04 × 25×10−9 / 1×8.854×10−12

Area = 1 x10−9 / 8.854 ×10−12

The area of this parallel plate capacitor by using the formula is 112.94 m2.

Conclusion

A capacitor is any device with two terminals that are meant to store any electrical energy in the capacitor and if the metal plates of this capacitor are arranged in parallel, they are termed a parallel plate capacitor. This parallel plate capacitor comprises insulating materials in between them such as vacuum, air, mica, glass etc. The formula for a parallel plate capacitor will be C=. The potential difference between two plates is directly proportional to the amount of electric charge stored in the plates of the capacitor. Moreover, the value of C is directly proportional to A (area) whereas it is inversely proportional to d (distance between the plates).