The Parallel Plate Capacitor MCQ Quiz - Objective Question with Answer for The Parallel Plate Capacitor - Download Free PDF

Correct option is : (3) Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates

Let the surface charge density be σ = q / A

Given dq/dt = constant

⇒ d/dt (q / A) = constant ⇒ (1 / A) × dq/dt = constant

It means displacement current is constant.

This system will act like a cylindrical wire.

The graph of magnetic field (B) vs radius (r) is:

Calculation:
For two capacitors in series, the total potential difference across both capacitors is equal to the battery voltage. The formula for the electrostatic energy stored in a capacitor is given by:

U = (1/2) × C × V²

Where:

• U is the energy stored in the capacitor,
• C is the capacitance,
• V is the voltage across the capacitor.

The capacitance of a parallel plate capacitor is given by:

C = (ε₀ × A) / d

Where:

• ε₀ is the permittivity of free space (for air, it is 1),
• A is the area of the plates,
• d is the separation between the plates.

For capacitor X, the dielectric constant is 1 (for air), and for capacitor Y, the dielectric constant is 2. The capacitance of Y will therefore be twice the capacitance of X. Since the capacitors are in series, the total capacitance is reduced, and the voltage across each capacitor is inversely proportional to their capacitance.

The energy stored in each capacitor depends on its capacitance, and since Y has twice the capacitance of X, the energy stored in X will be half of that stored in Y.

Correct Answer: Option 4 - 1 : 2

The correct answer is: 32 µF

CONCEPT:

• Capacitance in Series: When two capacitors are connected in series, the equivalent capacitance (C_eq) is given by the formula:

1 / C_eq = 1 / C₁ + 1 / C₂

CALCULATION:

Each capacitor originally has a capacitance of 40 µF.

One capacitor is filled with a dielectric material having a dielectric constant K = 4. Therefore, the new capacitance of this capacitor becomes:

C₁ = K × C = 4 × 40 µF = 160 µF

The second capacitor (C₂) remains unchanged at 40 µF.

Now applying the series combination formula:

1 / C_eq = 1 / 160 + 1 / 40

⇒ 1 / C_eq = (1 + 4) / 160 = 5 / 160

⇒ C_eq = 160 / 5 = 32 µF

Therefore, the correct answer is: 32 µF

Concept:

Two large, parallel, oppositely charged sheets (+σ on one, −σ on the other) can be treated as an ideal parallel-plate capacitor. A key property of such sheets, if they are “infinite” or large enough to neglect edge effects, is that:

The net electric field outside the region between the sheets is zero because the fields of the two sheets cancel there.

The net electric field inside the region between the sheets is nonzero (specifically, E = σ / ε₀ in a vacuum) and points from the positively charged sheet toward the negatively charged one.

Analysis:

At point P₁ (to the left of the +σ sheet), the net field is zero; hence the force on a test charge there is zero (F₁ = 0).

At point P₂ (between the two sheets), the net field is not zero; therefore a test charge experiences a nonzero force (F₂ ≠ 0).

At point P₃ (to the right of the −σ sheet), again the fields from the two sheets cancel, so the net field is zero and the force is zero (F₃ = 0).

Hence the forces are: F₁ = 0, F₂ ≠ 0, F₃ = 0.

Explanation:

Initially capacitor behave as a short circuit.

So current will be maximum.

Charge on capacitor will be zero.

Potential difference across capacitor will be zero.  

Concept:

A parallel plate capacitor consists of two large plane plates placed parallel to each other with a small separation between them. 

• The potential differences between the plates is, 
• \(V = \frac{Qd}{ϵ_0 A}\)
• Where Q = charge on the plate, d = distance between them, A = area of the plate. ϵ₀ is the permittivity of the space.

Explanation:

Let, the initial potential difference between the parallel plates is \(V = \frac{Qd}{ϵ_0 A}\)

When, the distance between parallel plates halved, d' = \(\frac d2\)

Then the final potential difference between the parallel plates is \(V' = \frac{Qd}{2ϵ_0 A}\)

V' = \(\frac V2\)

The potential difference between the parallel plates as the distance between parallel plates halved then the potential difference is decreased.

CONCEPT:

• Capacitance: The ability of an electric system to store an electric charge is known as capacitance.

C = Q/V

where Q is the charge on it, V is the voltage, and C is the capacitance of it.

• For the parallel plate capacitor, the capacitance is given by

\(C= \frac{KA\varepsilon _{0}}{d}\)

where A is the area of the plate, d is the distance between plates, K is the dielectric constant of material and ϵ is constant.

K = 1 for air or vacuum.

\(C= \frac{A\varepsilon _{0}}{d}\)

CALCULATION:

• Given that parallel plate capacitor

\(C= \frac{A\varepsilon _{0}}{d}\)

and C = Q/V

V = Q/C

V = Qd/(ε₀A)

So the correct answer is option 1.

CONCEPT:

• Capacitor: A capacitor is a device that stores electrical energy in an electric field.
• It is a passive electronic component with two terminals.
• The effect of a capacitor is known as capacitance.
• Capacitance: The capacitance is the capacity of the capacitor to store charge in it. Two conductors are separated by an insinuator (dielectric) and when an electric field is applied, electrical energy is stored in it as a charge.
  • The capacitance of a capacitor (C): The capacitance of a conductor is the ratio of charge (Q) to it by a rise in its potential (V), i.e.

  • C = Q/V

  • The unit of capacitance is the farad, (symbol F ).
  • Farad is a large unit so generally, we using μF.

EXPLANATION:

• The capacity of a conductor is defined as the ratio of charge given to the conductor to the rise in its potential i.e., 

\(\Rightarrow V = \frac{Q}{C}\)

∴ The capacity of a conductor depends on Q. It Also depends on the shape and size of the conductor as C = Aϵ_o/d. Therefore options 1 & 2 are correct.

CONCEPT:

• Capacitor: A capacitor is a device that stores electrical energy in an electric field. It is a passive electronic component with two terminals. The effect of a capacitor is known as capacitance.
• Capacitance: The capacitance is the capacity of the capacitor to store charge in it. Two conductors are separated by an insinuator (dielectric) and when an electric field is applied, electrical energy is stored in it as a charge.
  • Its unit is Farad, unit F. Farad is a large unit so generally, we using μF.

• The capacitance of a capacitor is C is given by;

\(\Rightarrow \text{C}=\frac{\text{Q}}{\text{V}}=\epsilon_o \frac{\text{A}}{\text{D}}\)

Where Q = charge, V = voltage, A = area of plate, D = distance between plate, ϵ = permittivity of medium. For air ϵ0 = 8.84 x 10-12 F/m.

• When the dielectric medium is K between the plates, then capacitance C is

\(\Rightarrow \text{C}=\frac{\text{Q}}{\text{V}}=\frac{\text{K }\!\!~\text{ }{{\epsilon}_{0\text{ }~\!\!\text{ }}}\text{A}}{\text{D}}\)

Where K is the dielectric constant of the medium, so ϵ = K ϵ0.

• After inserting the dielectric slab the capacitance is increased.

EXPLANATION:

• When the capacitor is charged by the battery and the battery is disconnected, then the charge on the capacitor remains the same. Therefore option 1 is correct.

Additional Information

• The capacitance of a capacitor is increased if the dielectric material is inserted between the plates of a capacitor.

\(\text{C}=\frac{\text{Q}}{\text{V}}=\frac{\text{K }\!\!~\text{ }{{\epsilon}_{0\text{ }~\!\!\text{ }}}\text{A}}{\text{D}}\), From this relation, we can say that-

• The capacitance is increased by the factor of K.

CONCEPT:

• The capacitance of a capacitor (C): The capacitance of a conductor is the ratio of charge (Q) to it by a rise in its potential (V), i.e.

C = Q/V

For a Parallel Plate Capacitor:

• A parallel plate capacitor consists of two large plane parallel conducting plates of area A and separated by a small distance d.

• Mathematical expression for the capacitance of the parallel plate capacitor is given by:

\(⇒ C = \frac{{{\epsilon_o}A}}{d}\)

Where C = capacitance, A = area of the two plates, ε = dielectric constant (simplified!), d = separation between the plates.

• The unit of capacitance is the farad, (symbol F ).

EXPLANATION:

Given - C₁ = 10 μF and d₁ = 2d

•  The capacitance of the parallel plate capacitor is given by:

\(⇒ 10 \, μ F = \frac{{{\epsilon_o}A}}{d}\)      ------- (1)

• The capacitance of the parallel plate capacitor when the distance between the plate is given by:

\(⇒ C_2 = \frac{{{\epsilon_o}A}}{d_1}= \frac{{{\epsilon_o}A}}{2d}\)      ------- (2)

On dividing equation 1 and 2, we get

\(⇒ \frac{10 \, μ F }{C_2}= \frac{\frac{{{\epsilon_o}A}}{d}}{\frac{{{\epsilon_o}A}}{2d}}=2\)

⇒ C₂ = 10/2 = 5 μF

CONCEPT: ​

• The capacitance of a capacitor (C): The capacity of a capacitor to store the electric charge is called capacitance.
  • The capacitance of a conductor is the ratio of charge (q) to it by a rise in its potential (V).

​

• The Capacitance of a Parallel Plate Capacitor with the distance between plates d and area of Cross-Section A is given as

 \(C = \frac{A\varepsilon _{0}}{d}\)--- (1)

ε0 is the Permittivity of free space.

• The energy (U) supplied by a battery to charge a Capacitor of Capacitance C at Potential Difference V is Given as:

U = CV²  ----- (2)

• The potential difference between two points at distance d and Electric Field E is Given as

V = E.d ---- (3)

CALCULATION:

The energy required to Charge a capacitor will be provided by Battery. 

So, Energy Required is Given by Eq (2)

Now, Putting Eq (1) and Eq (3) in Eq (2) We get

 \(U = \frac{A\varepsilon _{0}}{d} \times (Ed)^{2}\)

⇒ U = ε0E2 Ad

Additional Information

• The Energy Stored in a Capacitor is Given as \(\frac{1}{2}CV^{2}\) and Energy Provided by Battery is CV². So, to charge a capacitor twice of energy is to be supplied by the battery. 

CONCEPT:

• Capacitance: The capacitance tells that for a given voltage how much charge the device can store.

Q = CV

where Q is the charge in the capacitor, V is the voltage across the capacitor and C is the capacitance of it.

• When the capacitor is charged with the battery, conservation of energy can be applied here.
• Conservation of energy tells that if the potential energy of the battery decreases the energy of another part of the system must increase by the same amount.
• So, the energy from the battery is stored in the electric field between the plates.
• If no electric field existed between the plates, no energy would be stored between them.

EXPLANATION:

• A capacitor is a device used to store energy for a given voltage.
• When a battery is connected to the capacitor, the energy from the battery is stored in the capacitor.
• This energy is stored in the electric field between the plates.
• So the correct answer is option 4.

The correct answer is option 4) i.e. 1 : 3

CONCEPT:

• Capacitor: A capacitor is an electrical component with two terminals used to store charge in the form of an electrostatic field in it. 
  • It consists of two parallel plates each possessing equal and opposite charges, separated by a dielectric constant.
  • Capacitance is the ability of a capacitor to store charge in it.

The capacitance of a parallel plate capacitor is given by:

\(C =\frac{kAϵ_0}{d}\)

Where A is the area of the parallel plate, d is the distance between parallel plates, ϵ0 is the permittivity of free space ( 8.854 × 10-12 Fm-1) and k is the dielectric constant.

CALCULATION:

Given that:

Plate area = A 

Separation distance = d

The new separation distance, d' = d/3

Before: \(C =\frac{kAϵ_0}{d}\)

After: \(C' =\frac{kAϵ_0}{d'} = \frac{kAϵ_0}{d/3} =3 \frac{kAϵ_0}{d} = 3C\)

Ratio = \(\frac{C}{C'} = \frac{C}{3C} = \frac{1}{3}\)

Concept:

The RMS value of conduction current is \(I = \frac{V}{X_c}\)

where \(X_c = \frac{1}{ω C}\)

Therefore, I = VωC

Where ω = Frequency in rad/s, C = Capacitance of the capacitor, V = voltage supplied and X_c = Capacitive reactance

Calculation:

Given: C = 100 pF, V = 230 V, ω = 320 rad/s

⇒ I = 230 × 320 × 100× 10^-12 A

⇒ I = 7.36 × 10^-6 A

⇒ I = 7.36 μA

Hence the correct option is 7.36 μA. 

The correct answer is option 1) i.e. Capacitance increases, Voltage drop decreases

CONCEPT:

• Capacitor: A capacitor is an electrical component with two terminals used to store charge in the form of an electrostatic field in it. 
  • It consists of two parallel plates each possessing equal and opposite charges, separated by a dielectric constant.
  • Capacitance is the ability of a capacitor to store charge in it.
  • The capacitance of a parallel plate capacitor is given by:

\(C =\frac{kAϵ_0}{d}\)

Where A is the area of the parallel plate, d is the distance between parallel plates, ϵ0 is the permittivity of free space ( 8.854 × 10-12 Fm-1) and k is the dielectric constant.

• The capacitance C is related to the charge Q and voltage V across them as:

\(C = \frac{Q}{V}\)

EXPLANATION:

Capacitance, \(C =\frac{kAϵ_0}{d}\) 

• When a dielectric is placed in between the plates, the capacitance increases by a factor of dielectric constant k.
• When a capacitor is disconnected from the battery after being charged, it will retain the same charge. Therefore, Q remains constant.

From the relation \(C = \frac{Q}{V}\), Q is constant and C is increased.

⇒ V is decreased.

Thus capacitance increases, voltage drop decreases.