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## Karnataka 2nd PUC Physics Question Bank Chapter 2 Electrostatic Potential and Capacitance

### 2nd PUC Physics Electrostatic Potential and Capacitance One Marks Questions and Answers

Question 1.

Define potential energy of charge at a point.

Answer:

Potential energy of charge at a point is defined as the work done by the external force (equal and opposite to the electric force) in bringing the charge from infinity to that point.

Question 2.

Represent work done in bringing a test charge from one point to another in an electric field with the help of mathematical expression.

Answer:

W_{Rp} = work done in bringing a test charge from R to P = \(\int_{R}^{P} F_{e x t} \cdot d \vec{r}\)

Note : ff the given charge is positive and the test charge moved is also +ve then work done is -ve.) i.e., W_{Rp} = – \(\int_{R}^{p} \vec{F} \cdot d \vec{r}\)

Question 3.

Define potential energy difference between any two points.

Answer:

ΔU = U_{p} – U_{R} = W_{Rp}. The electric potential difference between any two points in an electric field is defined as the difference in the amount work done to transfer +1C of charge from infinity to those two points.

Question 4.

What is the reference potential energy taken at infinity?

Answer:

Zero. U_{∞} = 0

Question 5.

Give an expression for the electrical potential at a point due to a point charge.

Answer:

Question 6.

Give an expression for P.E. between any two points due to a given point charge.

Answer:

Question 7.

IIow does electric potential depend on V and 0 due to an electric dipole?

Answer:

V ∝ \(\frac{1}{r^{2}}\) and V ∝ cos θ, where jr is the distance of a point from the centre of a dipole and 0 is the angle between the line joining the point and the centre and the dipole axis.

Question 8.

Answer:

Question 9.

Give the expression for electric potential at a point on the axis of the dipole.

Answer:

Question 10.

Give the expression for electric potential at a point, due to a system of discrete point charges.

Answer:

Question 11.

What is the angle between the equipotential surface and electric field at that point?

Answer:

90° or \(\left(\frac{\pi}{2}\right)\) rad.

Question 12.

What is the shape of equipotential surface around the point charge?

Answer:

Spherical.

Question 13.

What does the term E = – \(\frac{d \mathbf{V}}{d x}\)signify?

Answer:

Electric field is in the direction along which the electric potential decreases sharply.

Question 14.

Express electrostatic potential energy in terms of electric potential.

Answer:

Question 15.

Give the expression for potential energy due to an electric dipole.

Answer:

Question 16.

Electric potential on the surface of a spherical shell is 1000V. What will be the electric potential at any point inside the shell?

Answer:

1000V. Electric potential at any point inside is the same as on the surface.

Question 17.

What is the electric field inside a conductor?

Answer:

Zero.

Question 18.

What is the direction of electric field intensity on a Gaussian surface around a point charge?

Answer:

Electric field intensity is normal to the surface enclosing a point charge.

Question 19.

What will be the electric field inside a cavity of any conductor?

Answer:

Zero. (All tuning circuits need protection from stray electric fields. Hence these circuits require electrostatic shielding. These are housed in the cavity or inside a metal box)

Question 20.

What is the effect of induced dipole moment on the external electric field?

Answer:

The induced dipole moment produces electric field which opposes the applied electric field.

Question 21.

Define electric polarization.

Answer:

The dipole moment per unit volume is called electric polarization.

Note : Electric polarization is a vector physical quantity \(\vec{P}\) = χ_{e} \(\vec{E}\), where χ_{e} is a constant characteristic of the dielectric.

Question 22.

What are linear isotropic dielectrics?

Answer:

Substances whose induced dipole moment act in the direction of the electric field are called linear isotropic dielectrics.

Question 23.

Define capacitance of a conductor.

Answer:

The ratio of electric charge on the conductor to its electric potential is known as the electrical capacitance of the conductor.

Question 24.

What is meant by an electrical capacitor? (July 2014)

Answer:

Electrical capacitor is an arrangement in which the capacitance of a conductor is increased by bringing an identical earthed conductor close and parallel to it.

Question 25.

How does electrical capacitance depend on the area of the plate?

Answer:

Capacitance is directly proportional to the area of the charged plate.

Question 26.

How does electrical capacitance of a parallel plate capacitor depend on the distance between the plates?

Answer:

Capacitance is inversely proportional to the distance between the plates or thickness of the dielectrics between the plates.

Question 27.

What is the net electric field intensity outside the plates of a capacitor?

Answer:

Zero.

Question 28.

What ¡s the net electric field intensity ¡n between the oppositely charged plates of a capacitor?

Answer:

Question 29.

Compare the capacitance of a parallel plate capacitor with and without the dielectric medium.

Answer:

C_{medium} = ε_{r}C_{air} Where ε_{r} represents the dielectric constant of the medium.

Question 30.

Write the equivalent capacitance of a number of identical capacitors connected in series.

Answer:

C_{s} = \(\frac{C}{n}\) where ‘n’ is the number of capacitors.

Question 31.

Write the equivalent capacitance of a number of identical capacitors connected in parallel.

Answer:

C_{p} = nC.

Question 32.

Give the expression for equivalent capacitance of a number of capacitors of different capacitances in series combination.

Answer:

Question 33.

Give the expression for equivalent capacitance of a number of capacitors of different capacitances in parallel combination.

Answer:

Question 34.

Give different expressions to find the energy stored in a capacitor.

Answer:

Question 35.

What is the amount of energy stored per unit volume in a capacitor.called?

Answer:

Energy density. Energy density = \(\frac { 1 }{ 2 }\)ε_{0}E^{2} where E – electric field between the plates.

Question 36.

What is Van de Graaff generator ?

Answer:

Van de Graaff generator is a machine/device to convert a low DC into a very high DC of the order 10^{6}V.

Question 37.

What is the value of electric potential due to a charge at its own location?

Answer:

The value of electric potential due to a charge at its own location is not defined. It is considered as infinite.

Question 38.

What is the amount of work done to move a point charge fron one point to another on an equipotential surface?

Answer:

Zero. This is because p.d = zero.

Question 39.

What is angle between the electric field at a point and the equipotential surface passing through the point?

Answer:

The angle between the electric field intensity \(\vec{E}\) and the equipotential surface is 90°.

Question 40.

What is a neutral point in a combined electric field?

Answer:

A neutral point is that point in a combined electric field at which the net field becomes zero.

Question 41.

A closed surface has an electric dipole. What flux will pass’ through the surface?

Answer:

Zero.

Question 42.

What will be electric potential at any point on the perpendicular bisector of an electric dipole?

Answer:

Zero.

Question 43.

What is the condition for a system of charges to be at equilibrium?

Answer:

For equilibrium, net potential energy of the system of charges should be zero.

Question 44.

Write the expression for work done by the force acting on an electric dipole to deflect it through a certain angle with respect to the uniform electric field.

Answer:

W = pE (1 – cos θ) whereis the dielectric dipole moment.

Question 45.

Define electrical capacitance of a conductor;

Answer:

Electrical capacitance of a conductor is defined as numerically equal to the quantity of charge required to raise the potential of the conductor by one unit.

Question 46.

Compare the electrical capacitance of a spherical capacitor with and without the presence of dielectric medium.

Answer:

C_{m} : C_{a} :: ε_{r} : 1 where ε_{r} is a dielectric constant.

Question 47.

Define SIU of electrical capacitance.

Answer:

If charge on a conductor is increased by one coulomb and electric potential is raised by 1 volt, then the electrical capacitance is said to be one farad (IF).

Question 48.

Define SIU of electric charge.

Answer:

SIU of electric charge is coulomb. If one ampere of current is maintained in a conductor for a period of one second then the net charge transferred is said to be 1 coulomb.

Question 49.

What is the effect of temperature on the dielectric constant of a dielectric medium?

Answer:

As the temperature is increased, the value of dielectric constant decreases.

Question 50.

Name the type of capacitors that are used as back up voltage sources for computers.

Answer:

Super capacitors. These are abbreviated as super caps.

Question 51.

Define one farad of electrical capacitance

Answer:

If one coulomb of charge is required to raise the potential of a conductor by one volt, then the capacitance of that conductor is said to be one farad.

Question 52.

What happens to the electrical capacitance of a capacitor, when a dielectric medium is introduced?

Answer:

The capacitance of a capacitor increases with the introduction of dielectric medium between the plates.

Question 53.

What is the effect of dielectric medium on the electric potential of a capacitor?

Answer:

The electric potential decreases with the introduction of dielectric medium.

Question 54.

What happens to the electrical capacitance of a conductor when it is brought closer to an earthed conductor?

Answer:

The capacitance of a charged conductor increases.

Question 55.

What is the direction of induced electric field in a dielectric medium?

Answer:

The induced electric field is opposite in direction to that of the applied electric field.

Question 56.

Write the unit of energy density?

Answer:

The SI unit of energy density is Jm^{-3}.

Question 57.

What is the a.c. resistance of the capacitor called?

Answer:

Capacitive reactance.

Question 58.

What is the value of capacitive reactance of a capacitor for a D.C. Voltage?

Answer:

Infinity. This is because frequency of D.C. is zero and capacitive reactance is inversely proportional to the frequency.

Question 59.

Define dielectric constant.

Answer:

Dielectric constant of a medium is defined as the ratio of absolute dielectric permittivity of a medium to that of a free space.

Question 60.

Give an example for a polar molecule.

Answer:

Hydrogen chloride. (Water, Ammonia, Carbon monoxide and Methanol are other examples for polar molecules).

Question 61.

Give an example for a non-polar molecule.

Answer:

Nitrogen. (Oxygen, Hydrogen, Benzene, Methane are other examples for non-polar molecules).

Question 62.

What is the magnitude of electric potential of the Earth?

Answer:

Zero. (The reference potential can be selected as zero inspite of the fact that electric potential of the Earth is negative).

### 2nd PUC Physics Electrostatic Potential and Capacitance Two Marks Questions and Answers

Question 1.

Define electric potential difference between any two points in an electric field and give the expression for electric potential difference.

Answer:

Electric potential difference between any two points in an electric field due to a point charge is defined as the difference in the electric potential energy acquired per unit positive test charge between these points.

Question 2.

Distinguish between polar and non-polar molecules.

Answer:

If the centers of positive and negative charges are separated by a distance, in the absence of an external electric field, then such molecules are called polar molecules.

,e.g.: HCl, H_{2}O.

If the centers of positive and negative charges coincide and get separated by a distance in the presence of an external electric field then they are called non-polar molecules.

e.g.: O_{2}, H_{2}.

Question 3.

Express electric potential in terms of charge density.

Answer:

Note: We know that electric potential at a point, due to a charged hollow conductor is given as,

Question 4.

Two Capacitors each of capacitances lOpF are connected in parallel. Calculate its capacitance.

Answer:

C_{p} = C_{1} + C_{2} = (10 + 10)μF = 20μF.

Question 5.

Calculate the equivalent capacitance between A and B of the following circuit, where C = 4 μF.

Answer:

C_{AB} = C = 4 μF. (Note : The givén circuit is a balanced network).

Question 6.

Mention any two factors on which the electrical capacitance of a parallel plate ; capacitor depends.

Answer:

Electrical capacitance of a parallel plate capacitor depends on

(a) area of a plate

(b) dielectric constant of the medium between the plates

(c) thickness of the dielectric medium.

Question 7.

What type of energy is stored in a capacitor? By what process can the energy stored in a capacitor be removed?

Answer:

A capacitor stores electrostatic energy. Energy can be removed by discharging the capacitor to the ground.

Question 8.

Show graphically : V α \(\frac{1}{r}\)

Answer:

Question 9.

Give the expression for electric potential energy due to (i) two point charges (ii) three non-collinear point charges.

Answer:

Question 10.

Give the expression for electric field intensity at the surface of a charged conductor with the explanation of the symbols used.

Answer:

\(\vec{E}\) = \(\frac{\sigma}{\varepsilon_{0}} \hat{n}\) where, n̂ is a unit vector normal to the surface, σ is the surface charge density and ε_{0} is the electrical permeability for free space.

Question 11.

Name the two factors on which the electric polarization depends.

Answer:

The electric polarization depends on,

- the dipole electric energy which tends to align the dipoles
- thermal energy which tends to disrupt (break) the alignment.

Question 12.

Write the formula for capacitance of a parallel plate capacitor.

Answer:

where, ε_{0} = 8.854 × 10^{-12} Fm^{-1} , ε_{r} is the dielectric constant for a given medium, ‘A’ is the area of cross-section and ‘d’ is the thickness of the dielectric medium.

Question 13.

If two capacitors are charged to potentials V, and V2 and there after connected in parallel, then what will be the potential difference in each of them?

Answer:

The net potential will be same in each of them. Common potential

### 2nd PUC Physics Electrostatic Potential and Capacitance Three Marks Questions and Answers

Question 1.

Draw a neat labelled diagram of Van de Graaff generator. Give the principle of its working.

Answer:

Principle : Spherical conductor holds the charge for a long period.

- The charge distribution will be uniform.
- Due to charge induction, the outer surface of the conductor will be charged.
- Discharge action of spiked bristles enables quick charging and discharging of the conveyor belt.

Working : MBI gets induced with a low positive voltage and due to induction, the belt gets positively charged. MB2 gets charged negative induction and delivers the positive charge on the surface of the conductor. The negative charges are sprayed back to the belt neutralizing it. A huge amount of charge can be stored by continuously working on the conveyor belt.

To get negative charge on the conductor, negative potential is supplied near the MBI. By working on two Van de Graaff generator, + 10^{6} V and – 10^{6} V, a p.d of 2 × 10^{6} V can be obtained. Van de Graaff generator converts low DC into high DC.

Question 2.

Give any three applications of Van de Graaff generator.

Answer:

Van de Graaff generator is used .

- to generate high DC voltages.
- to accelerate charged particles like electrons, protons and ions.
- in particle collisions.

Question 3.

Derive a relation between electric field and potential. (July 2014)

Answer:

Consider a point charge ‘+q’ Let ‘P’ be a point initially at infinity. Let ‘A’ be a point inside the . field region. Let ‘+1C’ be a unit positive charge moved from the point at infinity to point ‘A’. Let ‘A’ be at a distance of ‘r’ from the given point charge. Let ‘B’ be another point at distance of ‘dr’, from’A’towards the point charge+q.

The work done in moving positive test charge from A to B, against the force of repulsion is dW = -Fdr. The -ve sign indicates that the work is done against the direction of the force and dr is the displacement of +1C of test charge in the direction opposite to the electric field.

But electric potential dV = \(\frac{d W}{+q_{0}}\) and electric field E = \(\frac{\mathrm{F}}{+q_{0}}\)

When q_{0} = +1C then dW = dV, E = F

Hence dV = -Edr or E = –\(-\frac{d \mathrm{V}}{d r}\)

Question 4.

Give the expression for the iotential energy of a dipole in an external electric field with the help of a neat diagram.

Answer:

Potential energy = \(-\vec{p} \cdot \vec{E}\). Where \(\vec{p}\) – electric dipole moment \(\vec{E}\) – uniform electric field intensity.

Question 5.

Mention any three properties of equipotential surfaces. –

Answer:

- Equipotential surfaces around a point charge are concentric spherical surfaces and around a charged cylinder or line of charge are concentric cylindrical surfaces.
- Equipotential surfaces are planes which are parallel to the large charged plane.
- No work is said to be done when a positive test charge is moved from one point to another on an equipotential surface.
- An electric field line passing through a point on a Gaussian (equipotential) surface around a point charge, will be at right angles to the tangent drawn at that point.

Question 6.

Compare electrostatic potential at a point due to a point charge and an electric dipole.

Answer:

Electric potential at a point due to a point charge,

p-electric dipole moment, cosθ – cosine of the angle between the axis of the dipole and the line joining the point and the centre of the dipole and ‘r’ is the distance of the point from the Centre of the dipole.

Hence v ∝ cos θ and V ∝ \(\frac{1}{r^{2}}\)

Question 7.

Mention an expression for the electric potential energy of a system of two point : charges in the presence of an external field and explain the terms used.

Answer:

Electric potential energy of a system of two point charges,

Where,

q_{A}V_{A}– is the amount of work done in moving the point charge q_{A} from infinity to point A.

q_{B}V- is the amount of work done in moving the point charge ‘q_{B} from infinity to point B.

The term

represents the potential energy of the system due to an interaction between them.

Question 8.

Give the expression for electric potential at any point and at a point along the axis due to a short electric dipole.

Amswer:

Question 9.

Obtain an expression for electric potential due to a system of charges.

Answer:

Electric potential at P due to q_{1}, q_{2}, q_{3}, q_{4}…. q_{n}are given by

Since electric potential is a xcalar physical quantity

V = V_{1} + V_{2} + V_{3} + …… + V_{n} (scalar sum)

Note: V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}\), r ≥ R (for a charged spherical shell)

Question 10.

Mention any three results of electrostatics of conductors.

Answer:

- Electrostatic field is zero inside a conductor
- Electrostatic field must be normal to the surface at every point on the surface of a charged conductor .
- In a static situation, the interior of a conductor can have no excess of charge.
- Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface.

### 2nd PUC Physics Electrostatic Potential and Capacitance Five Marks Questions and Answers

Question 1.

How is the electric potential at a point due to a given charge measured? Obtain an expression for the electric. potential at a point due to an isolated point charge.

Answer:

The electric potential at a point due to a given point charge may be measured by finding the amount of work required to bring a unit positive test charge, from a point at infinity to that point inside a field region.

Let ‘P ’be a point at infinity. Let A, B and C be points inside the filed. Let BC = dx. Let the displacement be ‘dx ’ and ‘+ 1C’ is brought closer by ‘dx’ in a direction opposite to the direction of Electric filed.

Amount work done = – Fdx.

For ‘+ 1C’, of charge, F = E and dW = dV where ‘dV’ is electric potential defined as

Hence dV = -Edx. Electric potential at ‘A’

Hence electric potential at a point is inversely proportional to the distance of that point.

Question 2.

Derive an expression for the electric potential energy of a system of two point charges in the absence of an external electric field. (March 2014)

Answer:

Let q be the charge on a point charge placed at A. Let ‘+1C’ of test charge be placed at infinity.

It is assumed that +1C of charge is brought from infinity to the point ‘B’ at a distance of ‘r’ from ‘A’. The electric field is due to ‘q_{A}‘.

i.e., dV = -Edx for unit +ve test charge.

Now replacing ‘+1C’ by ‘q_{B}‘, ‘dV’ by dW and ‘E’ by F, we get

dW = -Fdx; dx – displacement between C & B.

Integrating both sides we get,

This amount of work done is stored in the form of electric potential energy.

Question 3.

Derive an expression for electric potential energy of a system of charges in an electric field.

Answer:

Consider point charges q_{1}, q_{2}, q_{3.} Let r_{12}, r_{23}, r_{31} be the distances between q_{1}, q_{2}, q_{3.} q_{2}, q_{3}; q_{3}, q_{1} respectively. The potential energy between q_{1} & q_{2}, will be

Total Energy of the system as a result of assembling the charges U = U_{1} + U_{2} + U_{3}.

In the presence of an external electric field the total electric potential energy of a system.

Where, Work done on q_{1} against the external field in bringing the charge from infinity to the distance r_{1} from the reference point = q_{1}V(r) and so on.

Question 4.

Obtain an expression for the equivalent capacitance of two capacitors connected in series. Write the expression for equivalent capacitance of ‘n’ number of unidentical capacitances of capacitors in series.

Answer:

Capacitors are said to be connected in series when

(i) same charge appears in each of them

(ii) voltage divides in the inverse ratio of their capacitances.

For ‘n’ unidentical capacitances,

Statement: The reciprocal of the equivalent capacitance of a number of capacitors connected in series is equal to the algebraic sum of the reciprocals of their individual capacitances.

Question 5.

Obtain an expression for the equivalent capacitance of two capacitors connected in arallel to a common voltage source.

Answer:

Capacitors arc said to be connected in parallel when

(i) the p.d. across each of these capacitors remains the same.

(ii) the net charge ‘Q’ divides in the direct ratio of their capacitances.

i.e., Q = Q_{1} + Q_{2} where Q = CV

i.e., CV = C_{1}V + C_{2}V

or C_{p} = C_{1} + C_{2}

For’n’ capacitors, C_{p} = C_{1} + C_{2} + C_{3} + ……… + C_{n}

For ‘n’ identical capacitors, C_{p} = nC

Statement: The equivalent capacitance of a number of capacitors connected in parallel is equal to the algebraic sum of their individual capacitances.

Question 6.

Derive an expression for the capacitance of a parallel plate capacitor with air as dielectric medium. (March 2015)

Answer:

Let ‘σ’ represent the surface density of the conductor ‘A’

Electric filed intensity at any point near a charged conductor is given by E = \(\frac{\sigma}{\varepsilon_{0}}\)

where, \(\frac{Q}{\sigma}\) = A and -‘A’ is the area of the charged conductor.

In general, C = \(\frac{\varepsilon_{0} \varepsilon_{r} A}{d}\) Where ε_{r} – dielectric constant of medium other than air.

Capacitance of a parallel plate capacitor depends directly on the area of cross section of the plate, dielectric medium between the two plates and inversely proportional to the thickness of the dielectric medium.

Question 7.

Derive an expression for the capacitance of a parallel plate capacitor with a dielectric medium other than air or vaccum.

Answer:

Let ‘σ’ represent the surface density of the conductor ‘A’.

Let ‘σ_{p}’ represent the surface density of charge in the dielectric due to polarisation of the dielectric by the electric field.

The net electric field due to the presence of dielectric becomes

The potential difference across the plates is

Where, d is theihickness of the dielectric medium.

Let, σ_{p} be proportional to the electric field.

The capacitance C with dielectric between the plates, is then

It is evident that C_{medium} = K times C_{air} and K = \(\frac{\varepsilon}{\varepsilon_{0}}>1\) for all dielectrics other than air or vaccum.

Question 8.

Obtain an expression for the energy stored in a capacitor.

Answer:

Let ‘C’ be the capacitance of a parallel plate capacitor. By definition C = \(\frac{Q}{V}\). By definition

amount of work done to raise the charge by ‘dq’ is dW = Vdq.

i.e., dW = \(\frac{q}{C}\) dq

where ‘q’ is the instantaneous value of charge on the conductor.

Total work done in charging the capacit0r is given by,

or W = \(\frac{1}{2} \frac{Q^{2}}{C}\), amount of work done is stored in the form of energy.

Using Q = CV, E = \(\frac { 1 }{ 2 }\)CV^{2}, where ‘V’ is the maximum voltage supplied to a capacitor.

Question 9.

Arrive at an the expression for electric potential at a point due to a dipole and hence write expression for electric potential at a point on the dipole axis.

Answer:

### 2nd PUC Physics Electrostatic Potential and Capacitance List of Formulae

### 2nd PUC Physics Electrostatic Potential and Capacitance Numericals and Solutions

Question 1.

A spherical hollow conductor is charged to – 250 V. If the radius of the conductor is 0.15 m, then calculate.

(i) Charge on the conductor.

(ii) Electric filed intensity on the surface

(iii) Electric field intensity at the centre.

(iv) Electric potential difference between A and B at distances of 0.20 m & 0.25m from the centre respectively.

Answer:

(iii) Electric field intensity at any point inside a charged hoi low-spherical conductor is zero.

Question 2.

Three point charges + 10nC, – 15nC and + 2QnC are placed at the corners A, B and C respectively of a square, ABCD of side, 0.1m. Calculate the amount of work done to transfer – 100nC from the point D to the point ‘O’.

Answer:

= 90[30 – 10.605]

= 1745.55 V

V_{0} = 1909.19 V

∴ p.d. = [V_{0} – V_{D}] = 1909.19 – 1745.55

Hence, W = (pd) (charge transferred)

= 163.64 × (- 100nC) = 163.64 × – 100 × 10^{-9}

= 1.636 × 10^{-5} J

Question 3.

Calculate the amount of work done by a uniform electric field of 200 Vm^{-1} on an electric dipole of length 0.05 m and charge 40nC in order that the angle between the axis and the field is equal to 180°. What is the amount of torque required to maintain that position?

Answer:

W = pE (1 – cos θ), p = (2l) q = (0.05) (40 × 10^{-9})

∴ p = 2 × 10^{-9} Cm

∴ W = (2 × 10^{-9}) (200) (1 – cos 180°) = 8 × 10^{-7} J

τ = pE sin θ

τ = 0 ∵ sin 180° = 0

Question 4.

Calculate the number of electrons to be transferred from a material body in order to charge it to + 5.5 nC.

Answer:

Q = nC

n = 34.37 × 10^{9} electrons.

Question 5.

The effective capacitances of two condensers are 3 µF and 16 µF, when they are connected in series and parallel respectively. Compute the capacitance of each condenser.

Answer:

Given C_{p} = 3 × 10^{-6}F.

C_{p} = 16 × 10^{-6}F.

hence C_{1}C_{2} = (3 × 10^{-6}) (16 × 10^{-6}) F^{2} = 48 × 10^{-12} F^{2}

∴ C_{1} – C_{2} = 8 × 10^{-6} …..(2)

Solving (1) and (2);

C_{1} + C_{2} = 16 × 10^{-6}

+ C_{1} – C_{2} = 8 × 10^{-6}

2C_{1} = 24 × 10^{-6}

∴ C_{1} = 12 × 10^{-6}F

and consequently C_{2} = 4 × 10^{-6} F

Hence the two capacitors are of capacitances 12 µF and 4 µF.

Question 6.

A parallel plate capacitor consists of two circular plates of radius 0.05 m each, which are separated by a distance of 1.2 × 10^{-3}m. Calculate its capacitance. If the difference between the two plates is reduced to half the initial value, then calculate its capacitance.

Answer:

Given : Radius of the plates = 0.05 m

Area of circular plates = πr^{2}

= 3.142 × 5 × 5 × (10^{-2})^{2} m^{2}

= 7.855 × 10^{-3}m^{2}

Thickness of the capacitor = 1.2 × 10^{-3}m

hence, C’= 2 × 57.96 × 10^{-12}F = 115.92 × 10^{12}F

C” = 115.9 pF

Question 7.

When two capacitors are connected in series across a 2 kV line, the energy stored in the system is \(5\left(\frac{5}{11}\right)\)J and when connected in parallel to the same voltage line, the energy stored in the system is (22) J. Find the capacitances of the individual capacitors.

Answer:

Given : E_{s} = \(\frac{60}{11}\)J, p = 2 × 10^{3} V

E_{p} = 22J, C_{1} = ? , C_{2} = ?

w.k.t E = \(\frac { 1 }{ 2 }\)CV^{2}

For a series combination C_{s} = \(\frac{C_{1} C_{2}}{C_{1}+C_{2}}\) and for a parallel combination C_{P} = C_{1} + C_{2}

Solving (2) and (3)

C_{1} + C_{2} = 11 × 10^{-6}

C_{1} – C_{2} = 1 × 10^{-6}

2C_{1} = 12 × 10^{-6}

C_{1} = 6 × 10^{-6} F

and hence C_{2} = (11 – 6) 10^{-6} F = 5 × 10^{-6} F

Therefore, the two capacitors are 5 µF and 6 µF.

Question 8.

Two capacitors 3 µF and 5µF are charged to 18 V and 25 V respectively. These are then connected in series. Calculate the net voltage and net charge. Find the loss of energy after connecting them in series.

Answer:

Given

C_{1} = 3 × 10^{-6}F, C2 = 5 × 10^{-6}F

V_{1} = 18V, V_{2} = 25 V

Q_{1} = C_{1}V_{1} = 3 × 10^{-6} × 18

Q_{1} = 54 × 10^{-6}C

Similarly Q_{2} = C_{2}V_{2} = 5 × 10^{-6} × 25

Q_{2}= 125 × 10^{-6}C

Total energy = E_{1} + E_{2} = 2048.5 × 10^{-6}J

Effective capacitance in series,

C_{s} = \(\frac { 15 }{ 8 }\) × 10^{-6} For C = 1.875 × 10^{-6}F

and V = V_{1} + V_{2} = 18 + 25 = 43V

∴ Net energy of the system after connecting in series,

E = \(\frac { 1 }{ 2 }\)CV^{2}

= \(\frac { 1 }{ 2 }\) × 1.875 × 10^{-6} × (43)^{2}

= 1.733 × 10^{-6} × 10^{-3}

= 1.733 × 10^{-6} J

Loss in the energy = (1.875 – 1.733) 10^{-6} J

= 0.142 × 10^{-6}J

i.e., ∆E = 1.42 × 10^{-5}J

Question 9.

Two capacitors 1 µF and 3µF are charged individually to voltages of 100 V and 200 V respectively. These are then connected in a parallel combination. Find the loss in the energy stored in the capacitors.

Answer:

Given: C_{1} = 1 µF , C_{2} = 3 µF

V_{1} = 100 V, V_{2} = 200 V

Q_{1} = C_{1}V_{1} = 1 × 10^{-6} × 100 = 10^{-4}

Q_{2} = C_{2}V_{2} = 3 × 10^{-6} × 200 = 6 × 10^{-4} c

Total energy before they are connected in parallel,

Question 10.

Three concentric metallic spheres A, B and C of radii lm, 2m and 3m respectively are charged to 0.56 nC, 2.12 nC and 5.0 nC. Calculate the potentials of the spheres.

Answer:

Question 11.

A parallel capacitor collects a charge of 3 µC when connected to a 1.5 V battery with air as dielectric. On replacing the air with a dielectric material the capacitor collects 9 µC of charge. Find the dielectric constant of the material and also the energy

stored in it with the material as dielectric.

solution:

Given: V = 1.5 V,Q_{1} = 3 × 10^{-6}C, Q_{2} = 9 × 10^{-6}C, E_{r} = ?

By definition, capacitance

Question 12.

You are provided with 3 capacitors, each of capacitance 2 pF. In how many possible ways can you connect these capacitors. Calculate the capacitance of each of the combinations.

Answer:

Question 13.

Two charges 5 × 10^{-8}C and -3 × 10^{-8}C are located 0.16 m apart. At what point(s) on the line joining the two charges is the electric potential zero? (July 2014)

Answer:

(i) For a point in between the line joining the charges,

V_{P} = V_{PA} + V_{PB} = 0

5x = 0.48 – 3x

or 8x = 0.48

∴ x = 0.06 m from the charge -3 × 10^{-8}C to its left.

(ii) For a point outside the linejoining the charges,

i.e., 5x = 0.48 + 3x

2x = 0.48; x = 0.24 m to the left of the charg -3 × 10^{-8}C

Question 14.

A spherical conductor of radius 0.12 m has a charge of 1.6 × 10^{-7}C distributed uniformly on its surface. What is the electric potential,

(a) inside the sphere?

(b) just outside the sphere?

(c) at a point 0.18m from the centre of the sphere?

Answer:

Given r = 0.12m, q = 1.6 × 10^{-7}C .

(a) Electric field intensity at any point inside the sphere is zero

(∵ Q_{inside} = 0)

but, electric potential at a point inside the charged sphere is the same as the potential on the surface.

(b) Electric potential on the surface orjust outside the surface

(C) Electric potential at a distance O.18m from the centre of the sphere,

Question 15.

A parallel plate capacitor with only air between the plates has a capacitance of 8 pF. What will be the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 6?

Answer:

Given C = 8 × 10^{-12} F, ε_{r} = k = 6, d’= d/2

i.e., C’ = 96 × 10^{-12} F = 96 pF

Question 16.

Three capacitors each of capacitance 9pF are connected in series.

(a) What is the total capacitance of the combination?

(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

Answer:

Question 17.

Three capacitors of capacitances 2pF, 3pF, and 4pF are connected in parallel.

(a) What is the total capacitance of the combination?

(b) Determine the charge on each capacitor, if the combination is connected to a 100 V supply.

Answer:

For a parallel combination of’n’ non-identical capacitors, their equivalent capacitance

C_{p} = C_{1} + C_{2} + C_{3} + …………. + C_{n}

Here C_{p} = (2 + 3 + 4)pF = 9pF

(b) Total charge Q = C_{p}V

ie., Q = 9 × 10^{-12} × 100 C

i.e., Q = 900pC

But Q_{1} : Q_{2} : Q_{3} : : C_{1} : C_{2} : C_{3}

Thus Q_{1} : Q_{2} : Q_{3} : : 2 : 3 : 4

and Q_{1} = \(\frac{2}{9}\) × 900pC = 200 pC

Q_{2} = \(\frac{3}{9}\) × 900pC = 300pC

Q_{3} = \(\frac{4}{9}\) × 900pC = 400pC

Question 18.

In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^{-3}m^{2} and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100V supply, what is the charge on each plate of the capacitor? (March 2014)

Answer:

Given A = 6 × 10^{-3}m^{2}, d = 3 × 10^{-3} m, V = 100V, C = ?,Q = ?

(a) We know that C = \(\frac{\varepsilon_{0} \mathrm{A}}{d}\)

(b) |Q| – CV= 17.708 × 10^{-12} × 100 = 1.77 × 10^{-9} C or |Q|= 1.77 nC.

Question 19.

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Answer:

Given C_{1} = 600 × 10^{-12} F, V_{1} = 200 V

but Q_{1} = C_{1}V_{1}

i.e, Q_{1} = 600 × 10^{-12} × 200 C

∴ Q_{1} = 12 × 10^{-8}C

Also given C_{2} = 600 × 10^{-12} F, Q_{2} = 0

Question 20.

Obtain the equivalent capacitance of the network in the figure given below. For a 300V supply, determine the charge and voltage across each capacitor.

Answer:

C_{s} = \(\frac{C_{2} C_{3}}{C_{2}+C_{3}}\) for a series combination of C_{2} and C_{3}

i.e., C_{s} = \(\frac{200 p F}{2}\) = 100 pF

Capacitors C_{1} and C_{s} are parallel to each other.

C_{p} = C_{1} + C_{s} = 100pF + 100pF = 200pF

The equivalent circuit is written as,

The equivalent capacitanoe of the circuit

Total charge, Q = C_{s}V = \(\frac{200}{3}\) × 10^{-12} × 300

i.e., Q = 2 × 10^{-8}C

The charges on C_{p} and C_{4} are the same i.e.,

i.e., Q_{p} = Q_{4} = 2 × 10^{-8}C

The charge 2 × 10^{-8} C branches out across the C_{1} C_{2} and C_{3} combination.

Q_{1} = Q_{2} = Q_{3} = 1 × 10^{-8}C and Q_{4} = 2 × 10^{-8}C

Question 21.

A parallel plate capacitor is to be designed with a voltage rating 1kV, using a material of dielectric constant 3 and dielectric strength about 10^{7} Vm^{-1}. For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

Answer:

Given V = 1 × 10^{3}V, ε_{r} = K = 3, E = 10^{7} Vm^{-1}.

Dielectric break down at 10% of E = \(\frac{10}{100}\) × 10^{7} Vm^{-1} × 10^{6} Vm^{-1} and C = 50 × 10^{-12} F.

i.e., the distance between the plates ‘d’ = 10^{-3} m

we àlso know that C = \(\frac{\varepsilon_{0} \varepsilon_{r} A}{d}\)

∴ Area of the plates A = \(\frac{C d}{\varepsilon_{0} \varepsilon_{r}}\)

Question 22.

A molecule of a substance has a permanent electric moment of magnitude 10^{-29} Cm. A mole of this substance is polarised by applying a strong electrostatic field of magnitude 106 Vm^{-1}. The direction of the field is suddenly changed by an angle of 60°. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample.

Answer:

Given, dipole moment of each molecule = 10^{-29} Cm

Total dipole moment of 1 mole of the substance = 6.023 × 10^{23} × 10^{-29} Cm

= 6.023 × 10^{-6} Cm

Initial potential energy, U_{i} = -pE cos θ = -6.023 × 10^{-6} × 10^{6} × cos 0° = -6.023J

Final P.E when, q = 60°, U_{f} = -pE cos Q = -6.023 × 10^{-6} × 10^{6} × cos 60°

= -3.0125 J

Change in potential energy = U_{f} – U_{i}

i.e, ∆P.E = [-3.0125 -( -6.023)]J = 3.0125 J

This energy loss by the substance is in the form of heat.

Question 23.

The area of a parallel plate capacitor is 6 × 10^{-3} m^{2}. The distance of separation of two plates is 3 × 10^{-3} m. Calculate the capacitance of the above capacitor. If the capacitor is charged to potential of 100V then calculate the charge on each plate. (March 2014)

Answer:

Given A = 6 × 10^{-3} m^{2}, d = 3 × 10^{-3} m. We know that C = \(\frac{\varepsilon_{0} A}{d}\)

The charge Q = CV = 17.7 × 10^{-12} m × 100 = 17.7 × 10^{-10}C = 1.77 nC.