Capacitance MCQ Quiz - Objective Question with Answer for Capacitance - Download Free PDF

Last updated on May 21, 2025

Latest Capacitance MCQ Objective Questions

Capacitance Question 1:

A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:

  1. zero at all places
  2. constant between the plates and zero outside the plates
  3. non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates
  4. zero between the plates and non-zero outside

Answer (Detailed Solution Below)

Option 3 : non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates

Capacitance Question 1 Detailed Solution

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:

1 (5)

Capacitance Question 2:

The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K and K2 with thickness \(\frac{3}{8}d\) and \(\frac{d}{2}\), respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates.
If K1 = 1.25 K2, the value of K1 is:

  1. 2.66
  2. 2.33
  3. 1.60
  4. 1.33

Answer (Detailed Solution Below)

Option 1 : 2.66

Capacitance Question 2 Detailed Solution

Calculation:
1 (3)

Ceq = (ε0 A) / (t1/K1 + t2/K2 + t3/K3)

Here C0 = ε0 A / d,   t1 = 3d / 8,   t2 = d / 2,   t3 = d / 8

K1 = K1,   K2 = K1 / 1.25,   K3 = 1

Given Ceq = 2C0

⇒ 2C0 = ε0 A / ( (3d / 8K1) + (d × 1.25 / 2K1) + (d / 8) )

⇒ 2ε0 A / d = ε0 A / ( (3d / 8K1) + (d / 2K1) + (d / 8) )

⇒ 2 = 1 / ( (3 / 8K1) + (5 / 8K1) + (1 / 8) )

⇒ K1 = 8 / 3 = 2.66

Capacitance Question 3:

Which of the following statements is INCORRECT with relation to the properties of dielectric material?

A) The electric field outside the dielectric gets modified due to the induced dipoles

B) The induced dipole aligns in the direction of applied electric field

C) Dielectrics contain free charges

D) Due to polarisation, dielectrics can store energy

  1. Only A and B 
  2. Only B
  3. Only C
  4. Only B and C

Answer (Detailed Solution Below)

Option 3 : Only C

Capacitance Question 3 Detailed Solution

Incorrect statement with relation to the properties of dielectric material dielectrics contain free charges.

Concept:

Dielectric

  • Dielectric is a material that does not have a free charge carrier for conduction. They however contain positive and negative charges which are bound together. 
  • They can be easily polarized which means charges inside the material will show a response to an electric field applied externally.
  • Dielectric materials are used to store energy. These materials exist in solid, liquid, and gaseous forms. Some examples of dielectric materials are:
  1. Solid Dielectrics – Plastic, Ceramic, Glass, and Mica
  2. Dielectric Liquid – Distilled Water.
  3. Dielectric Gas – vacuum, nitrogen, Dry Air,  and helium.

 

Explanation:

  • Polarizability- It is the tendency of an atom's electron cloud to be distorted from its normal shape by the application of an external electric field.

F7 Jayesh S 5-5-2021 Swati D3F7 Jayesh S 5-5-2021 Swati D4

  • It is denoted by \(\vec{p}\).
  • When we placed the Dielectric in an external field E, the positive and negative charges are displaced from their equilibrium position as shown in the above diagram. This process occurs throughout the volume of the dielectric.
  • This result in the formation of a large number of dipoles with each having some dipole moment along the direction of the electric field and
  • The whole phenomenon is called polarization.

Properties of Dielectric material

  • The band energy gap in the dielectric materials is very large.
  • The coefficient of temperature for resistance is negative.
  • They have high resistivity.
  • They have a strong attraction between electrons and the parent nucleus.
  • They have a very low electrical conductivity of these materials.
  • That's why they are basically electrical insulators.
  • The electric field outside the dielectric gets modified due to the induced dipoles
  • The induced dipole aligns in the direction of the applied electric field.
  • Due to polarization, dielectrics can store energy

Capacitance Question 4:

If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then

A. the charge stored in it, increases.

B. the energy stored in it, decreases.

C. its capacitance increases.

D. the ratio of charge to its potential remains the same.

E. the product of charge and voltage decreases.

Choose the most appropriate answer from the options given below:

  1. A, B and E only
  2. A, C only
  3. B, D and E only
  4. A, B and C only

Answer (Detailed Solution Below)

Option 2 : A, C only

Capacitance Question 4 Detailed Solution

Explanation:

A. The charge stored in it increases: This is correct because the capacitance increases and Q = CV with V constant.

B. The energy stored in it decreases: This is incorrect; the energy stored increases.

C. Its capacitance increases: This is correct because capacitance C = ε0 A / d increases as d decreases.

D. The ratio of charge to its potential remains the same: This is incorrect; the ratio Q / V is the same as capacitance, which increases.

E. The product of charge and voltage increases: This is incorrect as it equals the energy stored, which increases.

The correct option is 2) A, C only

Capacitance Question 5:

Four identical thin, square metal sheets, S1, S2, S3, and S4, each of side a are kept parallel to each other with equal distance d(<0 = ε0a2/d, where εis the permittivity of free space.

F1 sourav Teaching 14 11 24 D13

Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

List - I

List - II

(P)

The capacitance between S1 and S4, with S2 and S3 not connected, is

(1)

3C0

(Q)

The capacitance between S1 and S4, with Sshorted to S3, is

(2)

C0/2

(R)

The capacitance between S1 and S3, with S2 shorted to S4, is

(3)

C0/3

(S)

The capacitance between S1 and S2, with S3 shorted to S1, and S2 shorted to S4, is

(4)

2C0/3

 

 

(5)

2C0

 

  1. P → 3; Q → 2; R → 4; S → 5
  2. P → 2; Q → 3; R → 2; S → 1
  3. P → 3; Q → 2; R → 4; S → 1
  4. P → 3; Q → 2; R → 2; S → 5

Answer (Detailed Solution Below)

Option 3 : P → 3; Q → 2; R → 4; S → 1

Capacitance Question 5 Detailed Solution

Calculation:

For P

F1 sourav Teaching 14 11 24 D14

All are in series

\(C_{\text {eq }}=\frac{C_{0}}{3}\)

P → (3)

For Q

F1 sourav Teaching 14 11 24 D15

\(C_{e q}=\frac{C_{0}}{2}\)

→ (2)

For R

F1 sourav Teaching 14 11 24 D16

\(C_{\mathrm{eq}}=\frac{2 C}{3}\)

→ (4)

For S

F1 sourav Teaching 14 11 24 D17

⇒ Ceq = 3C0

→ (1)

∴ Correct answer is P → 3; Q → 2; R → 4; S → 1

Top Capacitance MCQ Objective Questions

What is the total capacitance in the given circuit?

F1 Shubham.B 19-01-21 Savita D1

  1. 7 F
  2. 13 F
  3. 4.3 F
  4. 16 F

Answer (Detailed Solution Below)

Option 2 : 13 F

Capacitance Question 6 Detailed Solution

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Concept:

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors' capacitances.

\(C_{{eq}}(parallel) = {C_1} + {C_2} + {C_3} + \ldots + {C_4}\)

When capacitors are connected in series, the total capacitance is less than the least capacitance connected in series.

\(\frac{1}{{{C_{eq}(series)}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots + \frac{1}{{{C_n}}}\)

Calculation:

For the two 6 F capacitors connected in parallel, the resultant capacitance will be:

Cnet = 6 + 6 = 12 F

Also, the two 2F capacitors connected in series connection will be equivalent to a single capacitor of value:

Cnet = 2/2 = 1 F

The resultant circuit is drawn as:

F2 Shubham Bhatt 3.3.21 Pallavi D8

Since the 1 F and 12 F capacitors are connected in parallel, the net capacitance will be:

Cnet = 1 + 12 = 13 F

What effect will be on the capacitance of a capacitor when the area of the parallel plate capacitor is decreased?

  1. It will increase
  2. It will decrease
  3. there will be no effect
  4. It will initially increase and then decrease

Answer (Detailed Solution Below)

Option 2 : It will decrease

Capacitance Question 7 Detailed Solution

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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

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

EXPLANATION:

Parallel Plate Capacitor:

F1 P.Y Madhu 13.04.20 D9

  • 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, εo = permittivity of free space and d = separation between the plates,

  • From the above equation, it is clear that the capacitance of the capacitor is directly proportional to the area of the parallel plate capacitor.
  • Hence, if the area of the parallel plate capacitor is decreased then the capacitance of the capacitor will decrease. Therefore option 2 is correct.

What happens to the potential difference between the capacitors parallel plates as the distance between parallel plates halved?

  1. Decreased
  2. Increased
  3. Remains constant
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : Decreased

Capacitance Question 8 Detailed Solution

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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.

The potential difference between the two plates of a parallel plate capacitor is _____________. (Q is magnitude of charge on each plate of area A separated by a distance d)

  1. Qd/(εoA)
  2. o/AQ
  3. Ad/(εoQ)
  4. QA/dεo

Answer (Detailed Solution Below)

Option 1 : Qd/(εoA)

Capacitance Question 9 Detailed Solution

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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.

In the inner region between the two charged plates of a parallel plate capacitor the electric field is equal to ____________. ('Q' is magnitude of charge on each plate of area 'A')

  1. εo / AQ
  2. Q / (εoA)
  3. A / (εoQ)
  4. QA / εo

Answer (Detailed Solution Below)

Option 2 : Q / (εoA)

Capacitance Question 10 Detailed Solution

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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

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

The electric field due to an infinite uniformly charged sheet is given by:

F1 P.Y Madhu 16.04.20 D3

\(E = \frac{σ }{{2{ϵ_0}}}\)

Where σ = surface charge density

EXPLANATION:

Parallel Plate Capacitor:

F1 P.Y Madhu 13.04.20 D9

  • parallel plate capacitor consists of two large plane parallel conducting plates of area A and separated by a small distance d.
  • It can be taken as two infinite charged sheet.

The electric field between them is given by:

F1 P.Y Madhu 16.04.20 D4

E = E1 + E2

\(E = \frac{σ }{{2{ϵ_0}}} - \left( {\frac{{ - σ }}{{2{ϵ_0}}}} \right) = \frac{{σ + σ }}{{2{ϵ_0}}} = \frac{{2σ }}{{2{ϵ_0}}} = \frac{σ }{{{ϵ_0}}}\)

Since σ = Q/A

So E = Q/(ϵ0 A)

Hence option 2 is correct. 

The potential to which a conductor is raised, depends on _______

  1. the amount of charge
  2. geometry and size of the conductor
  3. both (1) and (2)
  4. None of the above

Answer (Detailed Solution Below)

Option 3 : both (1) and (2)

Capacitance Question 11 Detailed Solution

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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 appliedelectrical 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 = Aϵo/d. Therefore options 1 & 2 are correct.

Which of the following options are incorrect? C is the capacitance of the capacitor, V is the voltage and Q is the charge of the capacitor.

  1. Energy stored in the capacitor is: 0.5 CV2
  2. Energy stored in the capacitor is: 0.5 QV

  3. Charge stored in the capacitor is: CV
  4. The capacitance equivalent when C1, C2, C3,….., Cn capacitors are connected in series: C1 + C2 + C3+ …….. + Cn

Answer (Detailed Solution Below)

Option 4 : The capacitance equivalent when C1, C2, C3,….., Cn capacitors are connected in series: C1 + C2 + C3+ …….. + Cn

Capacitance Question 12 Detailed Solution

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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).

C = Q/V

  • The work done in charging the capacitor is stored as its electrical potential energy.
  • The energy stored in the capacitor is

\(U = \;\frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}C{V^2} = \frac{1}{2}QV\)

Where Q = charge stored on the capacitor, U = energy stored in the capacitor, C = capacitance of the capacitor and V = Electric potential

Combination of capacitors:

  • Parallel combination: When two or more capacitors are connected in such a way that their ends are connected at the same two points and have an equal potential difference for all capacitor is called a parallel combination of the capacitor.

F1 P.Y Madhu 16.04.20 D1 2 1

  • Equivalent capacitance (Ceq) for parallel combination:

Ceq = C1 + C2 + C3

Where C1 is the capacitance of the first capacitor, C2 is the capacitance of the second capacitor and C3 is the capacitance of the third capacitor

  • Series combination: When two or more capacitors are connected end to end and have the same electric charge on each is called the series combination of the capacitor.

F1 P.Y Madhu 16.04.20 D1 1

  • Equivalent capacitance (Ceq) in series combination:

\(\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\)

EXPLANATION:

  • The energy stored in the capacitor is 0.5 C V2. So statement 1 is correct.
  • As the charge stored in the capacitor is given by Q = C V. Therefore the energy stored in the capacitor is 0.5 QV. So statement 2 is correct.
  • The charge stored in the capacitor is given by Q = C V. So statement 3 is correct.
  • When n capacitors are connected in series combination then equivalent capacitance is 
\(\Rightarrow \frac{1}{C_{net}}=(\frac{1}{{{C_1}}} + \;\frac{1}{{{C_2}}} + \;\frac{1}{{{C_3}}} + \ldots + \;\frac{1}{{{C_n}}})\)

If a dielectric is inserted between the parallel plate capacitor then capacitance will ______. 

  1. remains the same
  2. increase
  3. decrease
  4. increase initially and then decrease

Answer (Detailed Solution Below)

Option 2 : increase

Capacitance Question 13 Detailed Solution

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CONCEPT:

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 farad, (symbol F ).


Paralle Plate Capacitor:

F1 P.Y Madhu 13.04.20 D9

 

  • 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.

EXPLANATION:

  • When a dielectric slab of thickness t and dielectric constant K is inserted between the parallel plate capacitor, then the capacitance becomes

\(C = \frac{{{\epsilon_o}A}}{{d - t + \frac{t}{K}}}\)

  • From the above equation, it is clear that when a dielectric slab of thickness t is inserted then the effective distance between the parallel plate capacitor decreases, and hence capacitance increases. Therefore option 2 is correct.

Which one of the following statement is correct with regards to the material of electrical insulators?

  1. They contain no electron
  2. Electrons do not flow easily through them
  3. They are crystals
  4. They have more number of electrons than the protons of their surface

Answer (Detailed Solution Below)

Option 2 : Electrons do not flow easily through them

Capacitance Question 14 Detailed Solution

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CONCEPT:

  • Insulators: The substances through which electric charges cannot flow easily are called insulators.
  • In the atoms of such substances, electrons of the outer shell are tightly bound to the nucleus.
  • Due to the absence of free charge carriers, these substances offer high resistance to the flow of electricity through them.
  • Most of the non-metals like glass, diamond, porcelain, plastic, nylon, wood, mica, etc. are insulators. 

EXPLANATION:

  • Insulators possess high resistivity and low conductivity.
  • Their atoms have tightly bound electrons that do not move throughout the material.
  • Because the electrons are static and not freely roaming, a current cannot easily pass.

Railways Solution Improvement Satya 10 June Madhu(Dia)

  • An important difference between conductors and insulators is that when some charge is transferred to a conductor, it readily gets distributed over its entire surface.
  • On the other hand, if some charge is put on an insulator, it stays at the same place

The distance between the plates of a parallel plate capacitor is tripled keeping other parameters unchanged. The new capacitance will become-

  1. Three times of the initial capacitance
  2. One third of the initial capacitance
  3. Nine times of the initial capacitance
  4. Remains unchanged

Answer (Detailed Solution Below)

Option 2 : One third of the initial capacitance

Capacitance Question 15 Detailed Solution

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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

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

Parallel Plate Capacitor:

F1 P.Y Madhu 13.04.20 D9

  • 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, εo = permittivity of free space, and d = separation between the plates,

EXPLANATION:

Since \(C = \frac{{{\epsilon_o}A}}{d}\)

  • From the above equation, it is clear that the capacitance of the capacitor is inversely proportional to the distance between the plates of the parallel plate capacitor.
  • Hence, if the distance between the plates of a parallel plate capacitor is tripled then the capacitance of the capacitor will become one-third of the initial capacitance.
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