82. The capacitance of a parallel plate capacitor will get doubled if
(A) the area of each plate is doubled
(B) the area of each plate is halved
(C) the distance between the plates is doubled
(D) the distance between the plates is halved
Capacitance of a Parallel Plate Capacitor – Complete Theory, Formula, Derivation
Correct Answer
(A) and (D)
What is Capacitance?
Capacitance is the ability of a conductor or a capacitor to store electric charge. It is defined as the amount of charge stored on a conductor per unit potential difference across it.
Mathematically,
C = Q / V
where
- C = Capacitance
- Q = Charge stored on the capacitor
- V = Potential difference between the plates
The SI unit of capacitance is the Farad (F).
Formula for the Capacitance of a Parallel Plate Capacitor
For a parallel plate capacitor with air or vacuum between its plates, the capacitance is given by
C = ε0A / d
where
- C = Capacitance
- ε0 = Permittivity of free space
- A = Area of one plate
- d = Separation between the plates
This equation clearly shows that the capacitance depends directly on the area of the plates and inversely on the distance between them.
Understanding the Formula Physically
The larger the area of the plates, the more electric charge can be stored on them without causing a large increase in potential difference. Therefore, increasing the plate area increases the capacitance.
Similarly, when the distance between the plates is reduced, the electric field becomes stronger for the same amount of charge, allowing the capacitor to store more charge at the same potential difference. As a result, decreasing the plate separation increases the capacitance.
Thus, the capacitance depends on the geometry of the capacitor and the dielectric material present between the plates.
Detailed Option-Wise Analysis
Option (A): The Area of Each Plate is Doubled
From the capacitance formula,
C ∝ A
If the plate area becomes twice its original value,
A’ = 2A
Substituting into the formula,
C’ = ε₀(2A)/d
C’ = 2C
This means the capacitance becomes exactly twice its original value.
Physically, a larger plate area provides more surface over which electric charge can accumulate, thereby increasing the charge storage capacity of the capacitor.
Therefore, Option (A) is Correct.
Option (B): The Area of Each Plate is Halved
If the plate area is reduced to half,
A’ = A/2
Then,
C’ = ε₀(A/2)/d
C’ = C/2
The capacitance becomes half of its original value instead of doubling.
Reducing the plate area decreases the amount of charge that can be stored at a given potential difference.
Therefore, Option (B) is Incorrect.
Option (C): The Distance Between the Plates is Doubled
The capacitance is inversely proportional to the distance between the plates.
C ∝ 1/d
If the separation becomes twice its original value,
d’ = 2d
Then,
C’ = ε₀A/(2d)
C’ = C/2
Thus, the capacitance decreases to half of its original value.
Increasing the separation weakens the interaction between the plates, making it more difficult to store charge.
Therefore, Option (C) is Incorrect.
Option (D): The Distance Between the Plates is Halved
If the plate separation becomes half,
d’ = d/2
Substituting into the formula,
C’ = ε₀A/(d/2)
C’ = 2C
The capacitance becomes exactly twice its original value.
Reducing the separation allows the capacitor to store more charge for the same applied voltage because the electric field between the plates becomes stronger.
Therefore, Option (D) is Correct.
Summary of How Different Quantities Affect Capacitance
The capacitance increases when the plate area increases because more charge can be stored. It also increases when the distance between the plates decreases because the electric field becomes more effective at storing charge. If a dielectric material with relative permittivity greater than one is inserted between the plates, the capacitance also increases because the dielectric reduces the effective electric field inside the capacitor.
Real-Life Applications of Parallel Plate Capacitors
Parallel plate capacitors are widely used in electronic circuits for storing electrical energy, filtering signals, tuning radio frequencies, smoothing voltage fluctuations in power supplies, and timing applications. Modern touch screens, camera flash circuits, and many communication devices rely on the principles of capacitance discussed in this problem.
Exam-Oriented Key Concepts
Students should remember that the capacitance of a parallel plate capacitor depends only on the geometry of the plates and the dielectric medium. It does not depend on the charge stored or the applied voltage. The capacitance is directly proportional to the plate area and inversely proportional to the plate separation. Questions involving doubling or halving these quantities are among the most common conceptual questions in competitive examinations.
Final Answer
The capacitance of a parallel plate capacitor becomes twice its original value if either the area of each plate is doubled or the distance between the plates is reduced to half.
Correct Options: (A) and (D)
