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Test: Energy Stored in a Capacitor - JEE MCQ


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Test: Energy Stored in a Capacitor - Question 1

The charging of a capacitor through a resistance follows

Detailed Solution for Test: Energy Stored in a Capacitor - Question 1

When a battery is connected to a series resistor and capacitor, the initial current is high as the battery transports charge from one plate of the capacitor to the other.

The charging current exponentially approaches zero as the capacitor becomes charged up to the battery voltage.

The expression of charging current I, during process of charging is

The current and voltage of the capacitor during charging is shown below.

Test: Energy Stored in a Capacitor - Question 2

Capacitor A is charged by a 120 V battery and capacitor B is charged by 100 V battery. If the energy stored in both the capacitors is the same then find the ratio of the capacitance of capacitor A to capacitor B.

Detailed Solution for Test: Energy Stored in a Capacitor - Question 2

CONCEPT:

Capacitor:

  • The capacitor is a device in which electrical energy can be stored.
    • In a capacitor two conducting plates are connected parallel to each other and carrying charges of equal magnitudes and opposite signs and separated by an insulating medium.
    • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or a semi-conductor called a dielectric.

Capacitance:

  • The property to store energy in form of charge is called capacitance.
  • The unit of capacitance is Farad.
  • The charge on the capacitor (Q) is directly proportional to the potential difference (V) between the plates,

​⇒ Q ∝ V
⇒ Q =  CV
Where C = capacitance
Energy stored in the capacitor:

 

  • If a capacitor of capacitance C is charged by a potential difference V, then the energy stored in the capacitor is given as,

CALCULATION:
Given VA = 120 V, VB = 100 V, and UA = UB

 

  • We know that if a capacitor of capacitance C is charged by a potential difference V, then the energy stored in the capacitor is given as,


By equation 1 for the capacitor A,

By equation 1 for the capacitor B,

By equation 2 and equation 3,

Hence option 2 is correct.

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Test: Energy Stored in a Capacitor - Question 3

The work done in charging a capacitor is equal to? ('Q' is magnitude of charge on each plate of the capacitor between which the potential difference is 'V')

Detailed Solution for Test: Energy Stored in a Capacitor - Question 3

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 = ϵ0 A/d

Where ϵis the permittivity of free space, A is area and d is the distance between the plates.

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

Energy stored (U) in a parallel plate capacitor is given by:

U = (1/2)CV2 = Q2/2C

EXPLANATION:

Energy stored in a parallel plate capacitor (U) = Q2/2C.

Since C = Q/V

Work done (U) = Q2/2C = Q2/(2 Q/V) = QV/2

So option 3 is correct.

Test: Energy Stored in a Capacitor - Question 4

What is the energy density (energy stored per unit volume of space) of the parallel plate capacitor? (where E is the electric field between the two plates)

Detailed Solution for Test: Energy Stored in a Capacitor - Question 4

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 = ϵ0 A/d
Where ϵ0 is the permittivity of free space, A is area and d is the distance between the plates.
The electric field between the plates is given by:
E = σ/ϵ0 = Q/(ϵ0 A)
Where σ = Q/A = surface charge density

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

Energy stored (U) in a parallel plate capacitor is given by:
U = (1/2)CV2 = Q2/2C

EXPLANATION:

Since E = σ/ϵ0 = Q/(ϵA)

Q = E ϵA

Energy stored in a parallel plate capacitor (U) = Q2/2C

U = Q2/2C = (E ϵ0 A)2/2(ϵ0 A/d) = (εoE2Ad)/2

Energy density = U/Ad = εoE2/2

So option 4 is correct.

Test: Energy Stored in a Capacitor - Question 5

If the distance between the plates of a parallel plate capacitor is increased by maintain the constant charge on the capacitor, then the energy density between the plates will:

Detailed Solution for Test: Energy Stored in a Capacitor - Question 5

CONCEPT:

Capacitor:

  • The capacitor is a device in which electrical energy can be stored.
    • In a capacitor two conducting plates are connected parallel to each other and carrying charges of equal magnitudes and opposite sign and separated by an insulating medium.
    • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.​

​Parallel plate capacitor:

  • A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.
    • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.​

The energy stored in the capacitor is given as,


Energy density:

  • It is defined as the energy stored per unit volume of space between the plates.
  • Energy density u between the plates is given as,


Where C = capacitance of the capacitor, Q = charge on the plates, V = potential difference between the plates, and E = electric field intensity between the plates

EXPLANATION:

  • Energy density u between the plates is given as,

  • From the above equation, it is clear that the energy density of the parallel plate capacitors does not depend upon the distance between the plates of a parallel plate capacitor.
  • Therefore, if the distance between the plates of a parallel plate capacitor is increased by maintaining the constant charge on the capacitor, then the energy density between the plates will remain the same. Hence, option 3 is correct. 
Test: Energy Stored in a Capacitor - Question 6

A capacitor that stores energy of 0.5 J, and has capacitance of 1 μF, has potential difference of ______ across it.

Detailed Solution for Test: Energy Stored in a Capacitor - Question 6

Concept:

Energy stored in capacitor:

  • capacitor is a device to store energy.
  • The process of charging up a capacitor involves the transferring of electric charges from one plate to another.
  • 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 difference

Calculation:

Given: U = 0.5 J, C = 1 μF

V2 = 106
V = 1000 volt

Test: Energy Stored in a Capacitor - Question 7

The energy stored in a parallel plate capacitor is E and the potential difference between the plates is V, then the E is related to V as:

Detailed Solution for Test: Energy Stored in a Capacitor - Question 7

CONCEPT:

Capacitor:

  • The capacitor is a device in which electrical energy can be stored.
    • In a capacitor two conducting plates are connected parallel to each other and carrying charges of equal magnitudes and opposite sign and separated by an insulating medium.
    • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.​

​Parallel plate capacitor:

  • A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.
    • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.​

The energy stored in the capacitor is given as,


Energy density:

  • It is defined as the energy stored per unit volume of space between the plates.
  • Energy density u between the plates is given as,


Where C = capacitance of the capacitor, Q = charge on the plates, V = potential difference between the plates, and E = electric field intensity between the plates

EXPLANATION:

Given U = E

We know that the energy stored in the capacitor is given as,

Where C = capacitance of the capacitor, and V = potential difference between the plates

∵ U = E

For a parallel plate capacitor, the capacitance remains constant.
∴ E ∝ V2

Hence, option 2 is correct.

Test: Energy Stored in a Capacitor - Question 8

Two capacitors A and B of capacitance C and 2C are connected in series across a potential difference V. Find the ratio of the energy stored in capacitor A to capacitor B.

Detailed Solution for Test: Energy Stored in a Capacitor - Question 8

CONCEPT:

Capacitor:

The capacitor is a device in which electrical energy can be stored.

  • In a capacitor two conducting plates are connected parallel to each other and carrying charges of equal magnitudes and opposite signs and separated by an insulating medium.
  • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or a semi-conductor called a dielectric.


Capacitance:

  • The property to store energy in form of charge is called capacitance.
  • The unit of capacitance is Farad.
  • The charge on the capacitor (Q) is directly proportional to the potential difference (V) between the plates,

​⇒ Q ∝ V

⇒ Q =  CV

Where C = capacitance

Energy stored in the capacitor:

  • If a capacitor of capacitance C is charged by a potential difference V, then the energy stored in the capacitor is given as,

CALCULATION:

Given QA = QB = Q, CA = C, and CB = 2C

  • We know that when capacitors are connected in series, the magnitude of charge Q on each capacitor is the same.
  • We know that the energy stored in the capacitor is given as,


By equation 1 for the capacitor A,

By equation 1 for the capacitor B,

By equation 2 and equation 3,

Hence option 2 is correct.

Test: Energy Stored in a Capacitor - Question 9

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system

Detailed Solution for Test: Energy Stored in a Capacitor - Question 9

Concept:

Capacitor is defined as a component in an electrical circuit which has the ability to store energy in the form of an electrical charge producing a potential difference across its plates. It is like rechargeable battery.

Charge on capacitor is given by:

q = CV

where q is charge stored , V is potential difference across its plates, C is the capacitance of the capacitor.

Energy stored in capacitor is calculated using the formula, U:

where q is charge stored , V is potential difference across its plates, C is the capacitance of the capacitor.
Calculation:

Charge on capacitor

q = CV

When it is connected with another uncharged capacitor


Initial energy

Loss of energy  = Ui - Uf
i.e. decreases by a factor (2)

Test: Energy Stored in a Capacitor - Question 10

A device which is used in our TV set, computer, radio set for storing the electric charge is

Detailed Solution for Test: Energy Stored in a Capacitor - Question 10

Concept:

  • capacitor, a device for storing electrical energy, consisting of two conductors in close proximity and insulated from each other.
  • It is used to store the energy in presence of an electric field.
  • It is a device used to store an electric charge between two plates.
  • It consists of one or more pairs of conductors separated by an insulator.
  • The SI unit of capacitance is Farads (F).

Explanation:

A capacitor is used in our TV set, computer, and radio set for storing the electric charge.

Hence, the correct option is 3.

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