Capacitor with a Dielectric MCQ Quiz - Objective Question with Answer for Capacitor with a Dielectric - Download Free PDF

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

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.

Concept:

Capacitance of a Parallel Plate Capacitor:

• The capacitance C of a parallel plate capacitor with a dielectric material between the plates is given by the formula:
• C = (K × ε₀ × A) / d, where:
  • K = Dielectric constant of the material (dimensionless)
  • ε₀ = Permittivity of free space (8.85 × 10⁻¹² F/m)
  • A = Area of the plates (m²)
  • d = Distance between the plates (m)
• When a dielectric material (oil) is inserted between the plates, the capacitance increases by a factor of K (dielectric constant).
• If the oil is removed (K = 1 for air or vacuum), the capacitance will be reduced by a factor of K.

Calculation:

Initially, the capacitance is C = K × C₀, where C₀ is the capacitance without the oil.

Given that K = 2 with the oil, the capacitance becomes C = 2 × C₀.

When the oil is removed (K = 1), the capacitance reduces to:

C' = C / K = 2C / 2 = C / 2

∴ The capacitance of the capacitor becomes C / 2, which corresponds to Option 4.

Concept & Setup:

When two capacitors are connected in series, they each carry the same charge Q. Denote their capacitances by C₁ = 3 μF and C₂ = 2 μF. Let:

V₁ be the voltage across the 3 μF capacitor.

V₂ be the voltage across the 2 μF capacitor.

U₁ be the energy stored in the 3 μF capacitor.

U₂ be the energy stored in the 2 μF capacitor.

1) Voltage-Ratio Derivation:

Since the same charge Q appears on each capacitor in series, the voltage across each is

V₁ = Q / C₁,    V₂ = Q / C₂.

Hence,

V₁ / V₂ = (Q/C₁) / (Q/C₂) = C₂ / C₁.

With C₁ = 3 μF and C₂ = 2 μF, we get

V₁ / V₂ = 2/3.

2) Energy-Ratio Derivation:

The energy stored in each capacitor is

U = Q² / (2C)

(an equivalent formula is U = ½ C V², but using Q² / 2C is often simpler when the same charge flows). Therefore,

U₁ = Q² / (2 C₁)   and   U₂ = Q² / (2 C₂).

Thus, their ratio is

U₁ / U₂ = [Q² / (2C₁)] / [Q² / (2C₂)]  = C₂ / C₁.

Again, with C₁ = 3 μF and C₂ = 2 μF, we have

U₁ / U₂ = 2/3.

Putting It All Together:

V₁ / V₂ = 2/3 (the larger capacitance has the smaller voltage in series).

U₁ / U₂ = 2/3 (the larger capacitance also stores less energy when the charge is the same).

Often, textbooks may label the smaller capacitor as “C₁” and the bigger one as “C₂,” in which case the same formulas lead to a different symbolic ratio. But following the labeling in the statement (where V₁, U₁ refer to the 3 μF capacitor and V₂, U₂ refer to the 2 μF), the final results are:

V₁ : V₂ = 2 : 3

U₁ : U₂ = 2 : 3.

Calculation:

\(\frac{\mathrm{CdV}}{\mathrm{dt}}=\mathrm{I}_{\mathrm{d}}\)

\(\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\mathrm{I}_{\mathrm{d}}}{\mathrm{C}}\)

= \(\frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}}\)

Hence,

dV/dt = 100

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

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

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.

The correct answer is option 1) i.e. Q constant V and U 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 C is related to the charge Q and voltage V across them as:

C = \(\frac{Q}{V}\)

Capacitance is also 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 and k is the dielectric constant.

• Work done in moving a unit charge from one plate to another is done by voltage. Therefore work done = V.dQ = Energy stored in the capacitor.

WD = \(\int_0^QV.dQ = \int_0^Q \frac{Q}{C} dQ = \int_0^Q \frac{1}{C}[\frac{Q^2}{2}]_0^Q = \frac{1}{2}\frac{Q^2}{C}\)

EXPLANATION:

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

From C =\(\frac{kAϵ_0}{d}\), C increases by a factor of k

Therefore, from C = \(\frac{Q}{V}\), Q is constant and C is increased. ⇒ V is decreased.

​Energy = \(\frac{1}{2}\frac{Q^2}{C}\), Q is constant and C is increased. ⇒ Energy (U) is decreased.

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:

• From the above, it is clear that a capacitor (condenser) is used in an electrical circuit to store an electric charge. Therefore option 3 is correct.

NOTE:

• An electrometer is an electrical instrument for measuring electric charge or electrical potential difference.

Transformer:

• An electrical device that is used to transfer electrical energy from one electrical circuit to another is called a transformer.
  • The transformer which increases the potential is called a step-up transformer.
  • The transformer which decreases the potential is called a step-down transformer.

CONCEPT:

• A capacitor is a  device where two conductors are separated by an insulating medium that is used to store electrical energy or electrical charge
• The capacitance is defined as the ability to store charge or it is the number of charges stored per unit potential in a capacitor.
  • Capacitance is given by

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

Where Q = Charge and V = Potential difference

• The potential on a charged sphere is given by

\(\Rightarrow V = \frac{K Q}{R}\)

Where  K = Dielectric constant Q  = Charge, R  = Radius

CALCULATION:

Let r = radius after splitting, R = Radius before splitting

• The potential on a charged sphere is given by

\(\Rightarrow V = \frac{K Q}{R}\)

The above of the equation can be rewritten as

\(\Rightarrow \frac{Q}{V} = \frac{R}{K}\)

We know \(C=\frac{Q}{V}\),  then the above equation can be written as

\(\Rightarrow C = \frac{R}{K}\)

As it is given that, C = 1 μF, then

\(\Rightarrow \frac{R}{K} = 1 \ \mu F\)

After splitting,

\(\Rightarrow 8\times (\frac{4}{3}\pi r^3)=\frac{4}{3}\pi R^3\)

\(\Rightarrow r=\frac{R}{2}\)

• The potential of the smallest drop is given by

\(\Rightarrow V = \frac{r}{K}\)

Substituting the value of r in the above equation

\(\Rightarrow V = \frac{R}{2 K}\)

The substituting the value of \(\frac{R}{K} = 1 \ \mu F\) in  the above equation

\(\Rightarrow V = \frac{1}{2 }\mu F\)

Hence, option  1 is the answer

CONCEPT:

Dielectric constant

• Dielectric is a material that has poor electrical conductivity but has the capability to retain or store electric charge in it.
  • The dielectric constant of a given material is defined as the ratio of the permittivity of the material (ϵ) to the permittivity of the free space (ϵ0).
• The dielectric constant, 

\(\Rightarrow k = \frac{ϵ}{ϵ_0}\)

• The dielectric constant K of a material is defined as

\(⇒ K=\frac {\sigma}{\sigma - \sigma_p}\)

where σ is the induced charge density in the case of a conductor and σp is the induced charge density in the case of a dielectric in the same external electric field.

EXPLANATION:

• Since the dielectric constant K is 

\(⇒ K=\frac {σ}{σ - σ_p}\)

where σ is the induced charge density in the case of a conductor and σp is the induced charge density in the case of a dielectric in the same external electric field.

• σ and σ_p are positive
• σ > σ_p  
• Therefore K is always greater than 1. Therefore option 4 is correct

Additional Information

Conductors and dielectrics

• A conductor has free electrons.
• When a conductor is placed in an external electric field, these free electrons move and align themselves to create an induced charged density σ which produces an induced electric field such that the net field inside the conductor is zero. 
• Dielectrics are non-conducting substances. They do not contain free electrons.
• When a dielectric is placed in an external electric field, its molecules align themselves in such a way that charges are induced on the surfaces of the dielectric with having charge density σp, and thus an electric field is created inside the dielectric.
• This induced electric field Ein has a value less than that of the external electric field.
• Both these fields have opposite directions.
• Therefore the magnitude of the net electric field inside a dielectric is less than that of the external electric field.
• The direction of the net electric field inside the dielectric is the same as that of the external electric field.

CONCEPT:

Capacitor: 

• A capacitor is a two-terminal energy storage device that stores energy in the form of a static electric field during the positive half cycle and gives away during the negative half cycle of supply.

Parallel plate capacitor:

• When two parallel plates are connected across a battery, the plates are charged and an electric field is established between them, and this setup is known as the parallel plate capacitor.

​EXPLANATION:

The capacitance of a parallel plate capacitor: 

• The capacitance is the capacity of the capacitor to store charge in it.
• Two conductors are separated by an insulator (dielectric) and when an electric field is applied, electrical energy is stored in it as a charge.
• In capacitors, the metal plates are used to connect a voltage source between the dielectric medium. Hence, the electric charge is produced due to the presence of a dielectric medium and it is stored in that dielectric medium itself.

Concept:

When the capacitor discharges through the resistor, all the stored energy will be converted into heat. Thus, the heat produced in the resistor will be equal to the energy initially stored in the capacitor.

• 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 unit of capacitance is the farad, (symbol F).
• Energy stored (U) in the capacitor is given by:

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

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

Calculation:

Given,

C = 4 μF = 4 × 10-6 F

Voltage applied = 400 V

Energy Stored  = Heat energy produced through resistance.

\(U = \frac{1}{2}C{V^2} = \frac{1}{2} × {(4 × 10^{-6})} × {(400^2)}\)

⇒ Heat Produced = 32 × 10^-2 Joule

⇒ Heat Produced = 0.32 Joule.

CONCEPT: 

• A capacitor is an arrangement made to store electric charge and electrical energy, using two parallel metal plates that are separated by a dielectric medium.
• A capacitor is charged by applying a potential difference V between the metal plates.
• While charging charges will be stored on both metal plates
• Suppose two metal plates having charges q1 and q2 are arranged to form a capacitor having a capacitance C, then the potential difference between the plates is given by

\(V = \frac{q_{1} - q_{2}}{2C}\)

Where q₁,q₂ = The charges on the metal plates, V = Potential difference, C = Capacitance of the  capacitor

CALCULATION:

Let C = Capacitance of the capacitor, 

Given - q₁ = q, \(q_{2}= \frac{q}{2}\)

• The potential difference between the metal plates with different charges is given by

\(V = \frac{q_{1} - q_{2}}{2C}\)

Substituting the given values in the above equation it becomes 

\(\Rightarrow V = \frac{q- \frac{q}{2}}{2C} = \frac{q}{4C}\)

• Hence option 2 is the answer

Concept:

Energy stored in condenser:

A capacitor is a device which stores energy in the form of charge.

The process of charging up a capacitor involves the transferring of electric charges from one plate to another.

The energy stored in the capacitor is;

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

Q = charge stored on the capacitor 

U = energy stored in the capacitor 

C = capacitance of the capacitor  

V = Electric potential difference

Energy: The ability to do work is defined as Energy.

Power: It is defined as the ratio of work done to that of the time taken

Power = work / time

So, Energy = Power × Time  ----(2)

Calculation: 

Given:

C = 50μf, V = 3000V, t = 2 ms

From equation (1);

\(U=\frac{1}{2}\times50\times 10^{-6}\times(3000)^2\)

U = 225 J

From equation (2);

\(Power = \frac{225}{2\times 10^{-3}}\)

Power = 112.5 KilloWatt

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 electric field intensity at the outer region of the parallel plate capacitor is always zero whatever be the charge on the plate.
• The electric field intensity in the inner region between the plates of a parallel plate capacitor remains the same at every point.
• The electric field intensity in the inner region between the plates of a parallel plate capacitor is given as,

\(\Rightarrow E=\frac{σ}{\epsilon_o}=\frac{Q}{A\epsilon_o}\)​

• The potential difference between the plates is given as,

\(\Rightarrow V=\frac{Qd}{A\epsilon_o}\)

• The capacitance C of the parallel plate capacitor is given as,

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

Where A = area of the plates, d = distance between the plates, Q = charge on the plates, and σ = surface charge density

EXPLANATION:

• We know that the electric field intensity at the outer region of the parallel plate capacitor is always zero whatever be the charge on the plate.
• Therefore when the charge on the plates of a parallel plate capacitor is increased, the electric field intensity at the outer side of the plates will remain zero.