Capacitor with a Dielectric MCQ Quiz in मराठी - Objective Question with Answer for Capacitor with a Dielectric - मोफत PDF डाउनलोड करा

Concept:

• The device that stores electrical energy in an electric field is called a capacitor.
• The capacity of a capacitor to store electric charge is called capacitance.

Capacitor consists of two plates of a conductor and a dielectric insulator between them

\(C = \frac{{\epsilon A}}{d}\)

Where C = capacitance in farads, ϵ = Permittivity of dielectric, A = area of plate overlap in square meters, d = distance between plates in meters

Capacitance ∝ Area

If the area of the plates of the capacitor is reduced, the value of capacitance also reduces.

Capacitance of Parallel Plate Capacitor:

The capacitance of parallel plate capacitor depends upon

• The distance d between two plates
• The area A of medium between the plates

Explanation:

From above equation, it is clear that the capacitance of a capacitor is independent of the charge, Electric Field and voltage.

It is the geometry of the plates forming capacitor that changes the capacitance.

Hence, the correct option is geometry of the plates forming capacitor.

Concept:

Effect of Inserting a Dielectric Slab:

• When a dielectric slab is inserted between the plates of a charged capacitor, the following effects are observed:
  • The capacitance increases due to the dielectric constant of the material.
  • If the capacitor is isolated (not connected to a battery), the charge on the plates remains constant, as no external source is available to change the charge.
  • The potential difference between the plates decreases, as the dielectric reduces the effective electric field between the plates.
  • The stored energy in the capacitor decreases because energy is proportional to the square of the voltage, and the voltage decreases.
  • The electric field inside the capacitor also decreases, since the dielectric reduces the field between the plates.

Explanation:

Since the capacitor is isolated, the charge remains constant. The insertion of the dielectric only affects the electric field, voltage, and energy stored in the capacitor. The charge on the capacitor does not change because there is no external circuit to either supply or remove charge.

∴ The charge on the capacitor remains unchanged, which corresponds to Option 1.

Calculation: 

E between two plates is

⇒ \(\frac{\sigma}{\varepsilon_0}\)

and due to one plate is

⇒ \(\frac{\sigma}{2 \varepsilon_0}\)

so the force will be halved.

So new force F = 5 N.

∴ the force acting on the charged particle will become 5 N.

Concept: 

Capacitance of a Parallel Plate Capacitor: The capacitance C of a parallel plate capacitor is given by

C = \(\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}\)

Effect of Dielectric Slab: When a dielectric slab of thickness t and dielectric constant k is inserted between the plates, the capacitance changes to:

C^`= \(\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\left(0.2+\frac{\mathrm{d}}{\mathrm{k}}\right)}\)

Calculation: 

⇒\(\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}=\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{\left(0.2+\frac{\mathrm{d}}{\mathrm{k}}\right)}\)

⇒0.6 = 0.2 + \(\frac{0.6}{\mathrm{k}}\)

⇒k = \(\frac{3}{2}\) = 1.5

∴ The dielectric constant k of the slab is 1.5.

Parallel Plate Capacitor: It parallel plate capacitor consists of two large plane parallel conducting plates of area A and separated by a small distance d.

The mathematical expression for the capacitance of the parallel plate capacitor is given by:

\(C = \frac{{{ϵ_o}A}}{d}\)

C = capacitance

A = area of the two plates

εo = dielectric constant of free space

d = separation between the plates,

The charge is related ot the voltage as:

Q = CV, or

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

Analysis:

Let the initial voltage across the capacitor will be V₁.

\(V_1=\frac{Q}{C_1}\)   ---(1)

When the dielectric is introduced, the charge stored will remain the same, but the capacitance will change to:

\(C_2 = \frac{{{ϵ_m}A}}{d}\)

The new voltage stored now will be:

\(V_2=\frac{Q}{C_2}\)   ---(2)

After inserting a glass slal, the potential becomes 1/8 of the previous value, i.e.

\(V_2=\frac{V_1}{8}\)

Putting V₁ and V₂ from Equation (1) and (2), we get:

\(\frac{Q}{C_2}= \frac{1}{8}\times\frac{Q}{C_1}\)

C₂ = 8C₁

With \(C = \frac{{{ϵ_o}A}}{d}\), we get:

\(\frac{{{ϵ_mϵ_0}A}}{d}=8\times \frac{{{ϵ_0}A}}{d}\)

ϵ_m = 8

∴ The dielectric constant of the slab inserted will be 8.

Answer : 3

Solution :

We know, Q = C.V

∴ Q₁ = 10C and Q₂ = 5C

Energy stored, \(\mathrm{E}=\frac{1}{2} \mathrm{CV}^2\)

∴ \(\mathrm{E}_1=\frac{1}{2} \mathrm{C}_1 \mathrm{~V}^2=\frac{1}{2} \times 2 \mathrm{C} \times 25=25 \mathrm{~J}\)

Similarly, \(\mathrm{E}_2=\frac{1}{2} \mathrm{C}_2 \mathrm{~V}^2=\frac{1}{2} \times \mathrm{C} \times 25=12.5 \mathrm{~J}\)

∴ \(\frac{B_1-B_2}{Q_1-Q_2}=\frac{12.5}{5}=\frac{5}{2}\)

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 correct as it equals the energy stored, which increases.

∴ The correct option is 2) A, C and E only

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:

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

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:

• The capacitor is connected with a battery. If we introduce a dielectric between the plates of the capacitor, the charge on the plates and the capacitance will change.
• The potential across the plates is independent of anything but the emf of the battery. Therefore there will be no change in the potential difference across the plates. Therefore option 1 is correct.

​Additional Information

• The electric field between the plates, from Gauss law, is

\(⇒ E=\frac{\sigma }{ε_0}\)

• The capacitance is defined as

\(⇒ C= \frac {Q}{V} = \frac {\sigma A}{Ed} =\frac {ε_0 A}{d}\)

• Capacitance is independent of charge Q on the plates and potential difference V across the plates.
• If a dielectric of dielectric constant K is inserted between the plates of a parallel plate capacitor, the potential difference changes due to induced charges on the edges of the dielectric.

• The electric field inside the capacitor is 

\(⇒ E = \frac{\sigma - \sigma _p}{ε _0}\)

• The potential difference across the capacitor is 

\(⇒ V= Ed = \frac{\sigma - \sigma _p}{ε _0}d\)

• The ratio of the total electric field with dielectric 'E' and the total electric field without the dielectric 'E0' can be represented as

\(⇒ \frac {E}{E_0}= \frac{\sigma - \sigma _p}{\sigma} = \frac{1}{K}\)

• The capacitance of this capacitor is

\(⇒ C= \frac {Q}{V} = \frac {\sigma A}{Ed} =\frac {ε_0 \sigma A}{(\sigma-\sigma_p)d} = \frac {Kε_0A}{d}=KC_0\)

where C0 is the capacitance of the capacitor when there was no dielectric between the plates

CONCEPT:

Electric displacement

• Electric displacement D is defined as the charge per unit area that would be displaced across a layer of conductor placed across an electric field.
• Electric displacement can be mathematically represented as

⇒ D = ε₀E + P

where E is the electric field applied and P is the polarisation.

EXPLANATION:

• Since electric displacement D

⇒ D = ε₀E + P

• It is directly proportional to the electric field applied
• It is dependent on the polarisation
• Since polarisation depends on the dielectric constant, therefore D depends on the dielectric constant of the dielectric. Therefore option 1 is correct

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.

​Capacitance

• The charge on the capacitor (Q) is directly proportional to the potential difference (V) between the plates,

​⇒ Q ∝ V

⇒ Q =  CV

where C = capacitance

CALCULATION:

• We know that the charge can only flow when there is a potential difference.
• So the condition of not flowing the charge is,

⇒ V₁ = V₂

• Because when V₁ = V₂, the potential difference = 0, and then no charge will flow.
• Hence, option 3 is correct.