A parallel plate capacitor is a fundamental electrical component commonly studied in physics and electrical engineering, introduced in NCERT Class 12 Physics. The parallel plate capacitor is included in the chapter Electric Charges and Fields. It consists of two flat, parallel plates made of conductive material, such as metal, separated by a small distance. When a voltage is applied across these plates, a charge accumulates on the plates, creating an electric field between them.
Parallel Plate Capacitor
Here's a detailed explanation of the parallel plate capacitor:
Construction
The two plates are typically made of a conductive material like metal and are geometrically identical and parallel to each other. They are separated by a dielectric material or air, which is an insulator.
Working Principle
When a voltage (V) is applied across the plates, it creates an electric potential difference, causing electrons to flow from one plate to the other. This process continues until the charge on the plates reaches a steady state, and an electric field (E) is established between the plates.
Charge Accumulation
The positive charge accumulates on one plate, and an equal amount of negative charge accumulates on the other plate. This accumulation of charge creates an electric field between the plates.
Electric Field
The electric field strength (E) is directly proportional to the voltage (V) applied and inversely proportional to th distance (d) between the plates. The formula for electric field strength is E = V/d.
Capacitance (C)
The capacitance of a parallel plate capacitor is a measure of its ability to store electric charge for a given voltage. It depends on the area of the plates (A) and the permittivity of the dielectric material (ε), if present. The formula for capacitance is C = (ε * A) / d.
Dielectric Material
In many practical applications, a dielectric material is placed between the plates. The dielectric reduces the electric field strength between the plates, which, in turn, increases the capacitance. The permittivity of the dielectric material (ε) affects the amount by which the capacitance increases.
Applications
Parallel plate capacitors are used in various applications, including energy storage in electronic devices (such as power supplies and filters), signal processing (like in radios and amplifiers), and sensor elements in various measuring instruments.
Discharge
When the voltage source is disconnected, the capacitor can discharge over time as the stored charge flows back to the source. The rate of discharge depends on the resistance in the circuit.
In summary, a parallel plate capacitor is a basic electrical component that stores electrical energy in the form of an electric field between two conductive plates. Its capacitance depends on factors like plate area, plate separation, and the presence of a dielectric material. Understanding parallel plate capacitors is essential in the study of electrical circuits and electronics.
FAQs on Parallel Plate Capacitor
Q. A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
A
Capacitance between the parallel plates of the capacitor, C = 8 pF
Let us assume, initially, distance between the parallel plates was d and it was filled with air.
Dielectric constant of air, k = 1
Capacitance C is given by the formula,
C = = , since k = 1……………………..(1)
Where A = area of each plate
= permittivity of free space = 8.854
d = distance between two plates
If the distance between two plates is reduced to half, then distance between two plates =
Dielectric constant of a new substance, = 6
Then the resistance between two plates, = = ……………(2)
From equation (1) and (2) we get
= 12 pF = 96 pF
Q. An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A large number of 1 μF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
A . Requirements:
Capacitance, C = 2
Potential difference, V = 1 kV = 1000 V
Available:
Each capacitor, capacitance, = 1
Potential difference, = 400 V
Assumption:
A number of capacitors are connected in series and these series circuits are connected in parallel to each other. The potential difference across each row in parallel connection must be 1000 V and potential difference across each capacitor must be 400 V.
Hence, number of capacitors in each row is given as = 2.5
So the number of capacitor in each row = 3
The equivalent capacitance in each row is given as, = + = 3, so F
Let there be ‘n’ rows, each having 3 capacitors, which are connected in parallel
The total equivalent capacitance of the circuit is given as
= + + + + …….. upto n =
It is given = 2. So n = 6
So the final circuit will be 3 capacitors in each row and 6 rows connected in parallel, Total capacitors required = 3
Q. What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of μF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
A. Capacitance of the parallel capacitor, C = 2 F
Distance between two plates, d = 0.5 cm = 0.5 m
Capacitance of a parallel plate capacitor is given by the relation,
C = , where == 8.854
A = = = 1129 m = 1129 km
To avoid this situation, the capacitance of a capacitor is taken in F.
Q. The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply. (a) How much electrostatic energy is stored by the capacitor? (b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.
A.
A 2.26
Area of the parallel plate capacitor, A = 90 = 90
Distance between plate, d = 2.5 mm = 2.5 m
Potential difference across plates, V = 400 V
- Capacitance of the capacitor is given by, C =
where == 8.854
Electrostatic energy stored in the capacitor is given by the relation
= C = = = 2.55 J
- Volume of the given capacitor, V’ = A = 90 5
= 2.25
Energy stored is given by u = = = 0.113 J/
Also, u = = = = = , where E = electric intensity
News & Updates
Electromagnetic Induction Exam
Student Forum
Anything you would want to ask experts?Popular Courses After 12th
Exams: BHU UET | KUK Entrance Exam | JMI Entrance Exam
Bachelor of Design in Animation (BDes)
Exams: UCEED | NIFT Entrance Exam | NID Entrance Exam
BA LLB (Bachelor of Arts + Bachelor of Laws)
Exams: CLAT | AILET | LSAT India
Bachelor of Journalism & Mass Communication (BJMC)
Exams: LUACMAT | SRMHCAT | GD Goenka Test