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

Last updated on Jul 16, 2025

Latest Capacitance MCQ Objective Questions

Capacitance Question 1:

A circuit consists of two diodes, each having a forward resistance of 50 ohms and an infinite backward resistance. A 6V battery is connected across the circuit. Determine the current flowing through the 100 ohm resistor.

  1. Zero
  2. 0.02 A
  3.  0.03 A
  4. 0.036 A

Answer (Detailed Solution Below)

Option 2 : 0.02 A

Capacitance Question 1 Detailed Solution

Calculation:

The diode with reverse bias acts as an open circuit due to its infinite backward resistance, so no current flows through that branch.

The current flows only through the branch containing the forward-biased diode and the 100 ohm resistor.

Effective resistance of the path is: 50 Ω (diode) + 150 Ω + 100 Ω = 300 Ω

Using Ohm's law: I = V / R = 6 / 300 = 0.02 A

Answer: Option (B)

Capacitance Question 2:

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Two capacitors are charged separately and then connected in parallel with opposite polarities. Capacitor A has capacitance C and is charged to voltage V. Capacitor B has capacitance C and is charged to voltage V/2.After disconnecting the batteries, they are connected in such a way that the positive terminal of one is joined to the negative terminal of the other. What is the final energy stored in the configuration?

  1. 0
  2.  (1/8)CV2
  3.  (1/2)CV2
  4.  (1/16)CV2

Answer (Detailed Solution Below)

Option 4 :  (1/16)CV2

Capacitance Question 2 Detailed Solution

Calculation:

Initial charge on Capacitor A: q1 = CV

Initial charge on Capacitor B: q2 = CV/2

Since connected in opposite polarity, net charge: qtotal = q2 − q1 =- CV / 2

Let the final voltage across both be Vf. Then:

q1′ = C × Vf

q2′ = C × Vf

Total charge = CV / 2 = 2C × Vf → Vf = V / 4

Final energy:

U = (1/(2C))q12 / C + (1/(2C))q22  = (1/16)CV2

Correct Answer: Option 4

Capacitance Question 3:

Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is V. The potential difference between the points A and B (shown in the figure) is :

  1. 1 V

Answer (Detailed Solution Below)

Option 2 :

Capacitance Question 3 Detailed Solution

Calculation:

Potential difference between the plates = V, and the distance between them = 10 cm

⇒ Electric field E = V / 10

From the figure:

Vertical distance AC = 3 cm

Horizontal distance CB = 4 cm

So, AB forms a right triangle and is inclined at an angle.

Only the component of AB in the direction of electric field (i.e., horizontal) contributes to the potential difference.

⇒ VAB = E × horizontal component of AB

= (V / 10) × 4 = 4V / 10 = 2V / 5

Final Answer: 2V / 5

Hence, the correct option is (2).

Capacitance Question 4:

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : Net dipole moment of a polar linear isotropic dielectric substance is not zero even in the absence of an external electric field.

Reason (R) : In absence of an external electric field, the different permanent dipoles of a polar dielectric substance are oriented in random directions.

In the light of the above statements, choose the most appropriate answer from the options given below : 

  1. (A) is correct but (R) is not correct 
  2. Both (A) and (R) are correct but (R) is not the correct explanation of (A)
  3. Both (A) and (R) are correct and (R) is the correct explanation of (A)
  4. (A) is not correct but (R) is correct

Answer (Detailed Solution Below)

Option 4 : (A) is not correct but (R) is correct

Capacitance Question 4 Detailed Solution

Answer: (4)

Explanation:
In the absence of an external electric field (E), the dipole moments of polar dielectric molecules are randomly oriented due to thermal motion. Hence, the net polarization vector becomes zero.

Assertion (A): Since polar dielectrics are randomly oriented, Pnet = 0. ✔️ Correct.

Reason (R): If E is absent, polar dielectrics remain polar but their dipole directions are randomly distributed. ✔️ Correct.

Since both A and R are true, and R correctly explains A, the correct answer is (4).

Capacitance Question 5:

A 2 µF capacitor with 4 Ω resistor is connected as shown in the following circuit:

The current flowing through 2 Ω resistor is

  1. 9 amp
  2. 0.9 amp
  3. 3 amp
  4. 1.5 amp

Answer (Detailed Solution Below)

Option 4 : 1.5 amp

Capacitance Question 5 Detailed Solution

Calculation:

Calculate the equivalent resistance of the circuit.

The branch of 4 Ω will be open as there is capacitor which will act as open circuit due to capacitor.

The 2 Ω resistor (R2) is in parallel with the 3 Ω resistor (R1).

Total Resistance, R = (R1R2)/R1 + R2

⇒ R = 1.2 Ω

The net resistance is R = 1.2 + 2.8 = 4 Ω 

Calculate the total current flowing in the circuit using Ohm’s Law.

Using I = V / R:

⇒ I = 6 V / 4 Ω

⇒ I = 1.4 A

Top Capacitance MCQ Objective Questions

What is the total capacitance in the given circuit?

  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.

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

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:

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:

  • 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

 

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, 
  • 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 

When, the distance between parallel plates halved, d' = 

Then the final potential difference between the parallel plates is 

V' =

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

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.

CALCULATION:

  • Given that parallel plate capacitor

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:

Where σ = surface charge density

EXPLANATION:

Parallel Plate Capacitor:

  • 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:

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., 

∴ 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.

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 12 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:

 

  • 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

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

  • 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 13 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.

  • 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

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 14 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

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.

  • 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.

  • Equivalent capacitance (Ceq) in series combination:

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 

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:

  • 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:

Where C = capacitance, A = area of the two plates, εo = permittivity of free space, and d = separation between the plates,

EXPLANATION:

Since 

  • 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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