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

Calculation:

$\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}=\frac{\mathrm{it}}{\left(\frac{{\in }_{0}A}{\mathrm{d}}\right)}=\frac{\mathrm{itd}}{{\in }_0\left(\pi {\mathrm{r}}^2\right)}$

$\Rightarrow \mathrm{d}=\frac{{\in }_0\left(\pi {\mathrm{r}}^2\right)}{\mathrm{i}}\left(\frac{\mathrm{v}}{\mathrm{t}}\right)$

$=\frac{(9\times {10}^{-12})\left(\frac{22}{7}\right)(0.1{)}^2}{0.15}(7\times {10}^8)\mathrm{m}$

∴ d = 1320 μm

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:

Calculation:

C_eq = (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 C_eq = 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

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.

• It is denoted by $\overrightarrow{p}$.
• 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

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:

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

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

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

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

Calculation:

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

C_net = 6 + 6 = 12 F

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

C_net = 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:

C_net = 1 + 12 = 13 F

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

$C=\frac{{ϵ}_{o}A}{d}$

Where C = capacitance, A = area of the two plates, ε_o = 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.

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' = $\frac{d}{2}$

Then the final potential difference between the parallel plates is $V^{\prime }=\frac{Qd}{2{ϵ}_{0}A}$

V' = $\frac{V}{2}$

The potential difference between the parallel plates as the distance between parallel plates halved then the potential difference is decreased.

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

$C=\frac{KA{\epsilon }_0}{d}$

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.

$C=\frac{A{\epsilon }_0}{d}$

CALCULATION:

• Given that parallel plate capacitor

$C=\frac{A{\epsilon }_0}{d}$

and C = Q/V

V = Q/C

V = Qd/(ε₀A)

So the correct answer is option 1.

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:

$E=\frac{\sigma }{2{ϵ}_0}$

Where σ = surface charge density

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.
• It can be taken as two infinite charged sheet.

The electric field between them is given by:

E = E1 + E2

$E=\frac{\sigma }{2{ϵ}_0}-\left(\frac{-\sigma }{2{ϵ}_0}\right)=\frac{\sigma +\sigma }{2{ϵ}_0}=\frac{2\sigma }{2{ϵ}_0}=\frac{\sigma }{{ϵ}_0}$

Since σ = Q/A

So E = Q/(ϵ₀ A)

Hence option 2 is correct. 

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:

• 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 = Aϵ_o/d. Therefore options 1 & 2 are correct.

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.

• Equivalent capacitance (C_eq) for parallel combination:

C_eq = 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:

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

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 

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

$C=\frac{{ϵ}_{o}A}{d}$

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

$C=\frac{{ϵ}_{o}A}{d-t+\frac{t}{K}}$

• 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.

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. 

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:

• 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:

$C=\frac{{ϵ}_{o}A}{d}$

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

EXPLANATION:

Since $C=\frac{{ϵ}_{o}A}{d}$

• 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.