Plus Two Physics Chapter Wise Questions and Answers Chapter 2 Electric Potential and Capacitance is part of Plus Two Physics Chapter Wise Questions and Answers. Here we have given Plus Two Physics Chapter Wise Questions and Answers Chapter 2 Electric Potential and Capacitance.

Board |
SCERT, Kerala |

Text Book |
NCERT Based |

Class |
Plus Two |

Subject |
Physics Chapter Wise Questions |

Chapter |
Chapter 2 |

Chapter Name |
Electric Potential and Capacitance |

Category |
Plus Two Kerala |

## Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 2 Electric Potential and Capacitance

**Very Short Answer Type Questions (Score 1)**

Question 1.

A soap bubble is charged to a potential of 16V. Its radius is doubled. Then the potential of the bubble now become

a. 4 V

b. 2 V

c. 8 V

d. 0 V

Answer:

c. 8 V

Question 2.

Two combining spheres A and B of radius a and b respectively are at the same potential. The ratio of the surface charge densities of A and B is _____

Answer:

Question 3.

The small drop of the same size are charged to V volt If n such drops to form a large drop, its potential will be

a. Vn^{2/3}

b. Vn

c. V/n

d. V/n^{2}

Answer:

Question 4.

Observe the relationship between the first two terms and fill in the blanks.

Polar: H_{2}O

Non-Polar: ____

Answer:

Question 5.

When capacitors are connected in series the effective capacitance

a. zero

b. increases

c. constant

d. decreases

Answer:

d. decreases

Question 6.

“An electric field defined as the path along which a test charge would move if it were free to do so.” State whether this statement is true or false?

Answer:

False. Electric field due to a given charge is defined as the space around the charge in which electrostatic force is left.

Question 7.

If there are n capacitors in parallel connected to a V volt source, then the energy stored is equal to

Answer:

Question 8.

Example of a nonpolar molecule is

a. HCI

b. O_{2}

c. H_{2}O

d. NH_{3}

Answer:

b. O_{2}

Question 9.

The dimensional formula of capacitance is

Answer:

Question 10.

A sheet of aluminum foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

Answer:

Remain unchanged

Question 11.

“The network done by a conservative force on an object moving along a closed loop is zero”. Check whether this statement is true or false?

Answer:

True.

Question 12.

Observe the relationship between the first two terms and fill in the blanks.

A capacitance of an isolated spherical conductor:

A capacitance of the parallel plate capacitor: _______

Answer:

Question 13.

A parallel plate capacitor is made by stacking equally spaced plates connected alternately. It the capacitance between any two adjacent plates is C, then the resultant capacitance is _____

Answer:

(n – 1) C.

Question 14.

At a point in space, the electric field points toward the north. In the region surrounding this point, the rate of change of potential will be zero along

Answer:

east-west

Question 15.

“An unchanged conductor A is brought near a positively charged conductor B. Then the charge on B will increase, but the potential of B will not change”. Check whether this statement is true or false?

Answer:

False. An unchanged conductor A is brought near a positively charged conductor B. Then the charge on B will not change, but the potential of B will decrease.

Question 16.

The work done in taking a unit positive charge from P to A is W_{A} and from P to B is W_{B
}

Answer:

c. W_{A }= W_{B}

Question 17.

“ It V is zero at a certain point, then E should be zero at that point”. Check whether this statement is true or false?

Answer:

False. It V is zero at a certain point, then E may or may not be zero.

Question 18.

Variation in potential is maximum if one goes

a. along the line of force

b. perpendicular to the line of force

c. in any direction

d. none of these

Answer:

a. along the line of force

Question 19.

For the configuration of media of permittivities ε_{0}, ε, and ε_{0} between parallel plates each of area A, as shown in the figure, the equivalent capacitance is

Answer:

Question 20.

The electric field lines are closer together near object A than they are near the object B. We can conclude that

a. the potential near A is greater than the potential near B

b. the potential near A is less than the potential near B

c. the potential near A is equal to the potential near B

d. nothing about the relative potentials near A and B.

Answer:

d. nothing about the relative potentials near A and B.

Question 21.

In the electric field of a point charge q, a certain charge is carried from point A to B, C, D and E. Then the work done

a. is least along the path AB

b. is least along the path AD

c. is zero along all the paths AB, AC, AD and AE

d. is least along AE

Answer:

(c) ABCDE is an equipotential surface, on an equipotential surface, no work is done in shifting a charge from one place to another.

Question 22.

Observe the relationship between the first two terms and fill in the blanks.

Series combination :

Parallel combination: _____

Answer:

**Short Answer Type Questions (Score 2)**

Question 1.

Two spheres of Copper of same radii, one hollow and the other solid are charged to the same potential. Which sphere has more charge? 2

Answer:

Both would carry equal charge as capacity depends on a radius.

Question 2.

In a series combination of capacitors, charge on each capacitor is the same but the potential is different

a. In series combination, the equivalent capacitance is always than any of the individual capacitance. 1

b. The effective capacitance of n capacitors be connected in parallel is ____

Answer:

a. less

Question 3.

Van de graft generator is a device which can develop very high electrical potential.

a. Explain the working principle of the above device. 1

b. Is it possible to obtain negative as well as positive potential using the above device? Justify. 1

Answer:

a.

1.The action of sharp points, ie, the phenomenon of corona discharge.

2. The charge given to a hollow conductor is transferred to the outer surface and is uniformly distributed over it.

b. Yes. If a +ve charge is given +ve potential is obtained and if ve charges is given ve potential is obtained.

Question 4.

Study the figure (i) and (ii). Find the potential at A and B. 2

Answer:

- A potential at A is positive since A is nearer to the positive charge.
- A potential at B is negative since B is nearer to the negative charge.

Question 5.

A potential difference is field property. Potential energy is the interaction energy of charge configuration.

a. if they are not the same, then define them. 1

b. Also, state how they are related. 1

Answer:

a. The electrostatic potential at a point in an electric field is defined as the work done in bringing a unit positive charge from infinity to that point against the field. An electric potential difference between two points is equal to the work done in moving a unit test charge between them.

The potential energy of a system of charges is defined as the work done in bringing the charges from infinity to form the system

b. Potential energy, U = Potential difference x Charge

Question 6.

Two copper sphere of radii, one hollow and the other solid is charged to the same potential.

a. Which of the two will hold more charge? 1

b. What is your justification? 1

Answer:

a. Both will hold the same charge

b. The capacity is small

Question 7.

The dotted circles indicate the surface such that they lie equidistant from the charge +q.

a. What is the name of these surfaces? 1

b. Write any two properties of these surfaces. 1

Answer:

a. Equipotential surfaces.

b.

i. No work is done to move a charge from one point to another along the equipotential surface,

ii. The direction of the electric field is always perpendicular to the equipotential surface.

Question 8.

One eV is the kinetic energy acquired by an electron accelerated on an electric field of p.d.one volt

a. How are lev and 1Mev related to joule? 1

b. Write the dimensional formula of electrostatic potential? 1

Answer:

a. 1 MeV= 10^{6 }eV= 1.6 × 10^{-13}J

1 eV = 1.6 × 10^{-19}J

b.

Question 9.

Potential of conduction changes on charging.

a. How is a charge related to the potential? 1

b. What is the ratio called? 1

Answer:

a. On charging a body its potential rises.

The potential of the body is proportional to the quantity of charge.

it is called capacity.

Question 10.

Choose the correct alternative from the bracket

a. The unit of capacitance is (Farad/Henry)

b. The net charge on a charged capacitor is ...........

Answer:

a. Farad

b. Zero

**Short Answer Type Questions (Score 3)**

Question 1.

Consider a parallel plate air capacitor of plate area of 100 cm^{2} and separation 4 mm. A potential difference of 200 V is established between its plates by a battery.

a. Calculate the capacitance and charge on the capacitor. 1

b. After disconnecting the battery, the space between the plates is filled by ebonite (k = 2.6). Then calculate the capacitance and charge on a capacitor. 1

c. Compare the surface-charge density on the plates in the cases (a) and (b). 1

Answer:

Q remains the same.

c. Since the charge remains unchanged, surface density in two cases is the same.

Question 2.

a. It is not possible to make a spherical conductor of capacity 1 Farad. Explain. 1

The figure shows the variation of voltage V across the plates of two capacitors A and B versus charge ‘Q’ stored on them. Which of the two capacitors has higher capacitance? Give reasons for your answer. 2

Answer:

a. The capacitance of a spherical conductor

∴ The radius of a spherical conductor

r = 9 × 10^{6} km, which is not possible.

From the graph, it is clear that V_{A} < V_{B}

∴ It follows that C_{A} > C_{B} i.e., capacitor A has a higher capacitance.

Question 3.

The figure below represents the electric field lines in a region.

a. If an electron is placed at A, in which direction will it move?

b. The nature of electric field lines in the region ACBDA

i. Uniform field

ii. Non-Uniform field

iii. Radical field

iv. Sinusoidal field 1

c. A charge ‘q’ is carried from A to B through two different paths ACB and ADB. In which path more work is done? Justify. 1

Answer:

a. Towards B

b. i. Uniform field

c. Same An electric field is a conservative field. Hence work done is independent of path.

Question 4.

In electrostatics, both the electric field intensity (E) and potential (V) are closely related.

a. What do you understand by the term electrostatic potential? 1

b. Show that the line integral of the electric field along any closed path is zero. 2

Answer:

a. The electrostatic potential at a point is defined as the work done by an external agent in bringing a unit +ve charge from infinity to that point against the electrostatic force of the field.

b: Consider a close path ACBDA in an electric field.

The line integral of an electric field along the closed path ACBDA is,

Question 5.

An electric potential difference is a work done to bring a unit +ve test charge from one point to another in an electric field.

a. What is the relation between the electric field and electric potential ? 1

b. Can electric potential be necessarily zero at a point where the electric field strength is zero? Explain. 1

c. Potential at a point is given by the equation V = 3x^{2} + 2y^{2} + 4z^{2}. Calculate the magnitude of field components at (2,1,1). 1

Answer:

a. E= dv/dr

b. No. E = dv/dr when E = 0, then V is a constant.

c.

Ex = -6x = -12 N/C,

Ey = -4y = -4 N/C

Ez= -8z = -8 N/C

Question 6.

Match the following

Answer:

Question 7.

a. In the case of a parallel plate capacitor, its capacity with vacuum or air as the dielectric is .............. 1

b. Derive the expression for the energy density of a parallel plate capacitor? 2

Answer:

a.

Question 8.

Fill in the blanks.

The substances which are poor conductors of electricity are known as............... There are two types of........... and ............ In the presence of an electric field ................ acquire an induced dipole moment. 3

Answer:

Dielectric, polar and nonpolar dielectric, non-polar dielectric.

Question 9.

An arrangement of conductor used to increase the capacitance of a conductor is called a capacitor. There should be a minimum of two conductors to form a capacitor.

a. Find the effective capacity of the combination shown below. 1

b. Three capacitors 2 μ F, 5 μ F, and 10 μ F are joined in (a) series,

(b) parallel. Find the equivalent capacitance. 2

Answer:

a. Effective capacity of the three 2 μ F capacitors in parallel is 2 + 2 + 2 = 6 μ F

Effective capacity of 2 μ F and 6 μ F in series

is

Question 10.

There are one copper sphere and a copper shell with radii 5 cm and separated by a distance 30 cm from the centre. They are charged to a potential of 10 V on their surfaces. The solid sphere attained a charge of G.

a. What is the potential at the centres of sphere and shell? 1

b. How much is the charge gained by the shell? 1

c. If we move an electron around the shell and around the solid sphere in a circular path, then what is the work done for that motion? 1

Answer:

a. 10 V. Because the potential is the same at side and surface of spherical conducting bodies.

b. The same charge of a sphere, i.e., Q (dimensions are the same).

c. Zero, because it is an equipotential surface.

Question 11.

The generation of emf when the magnetic flux associated with coil changes is known as electromagnetic induction.

a. Mention the factors on which the self-inductance of a solenoid depends. 1

b. Calculate the energy stored in an inductor of inductance 50 mH when a current of 2 A is passing through it. 1

c. Two identical loops one of copper and other of aluminum are rotated with the same speed in the same magnetic field. In which case the

i. induced emf

ii. induced current will be more and why? 1

Answer:

a. Self inductance,

It depends on the length of the coil I, Area A, number of turns n and permeability .

b.

c.

i. induced emf will be the same because the rate of change of flux linkage is the same for both loops.

ii. Induced current will be more in copper because the resistivity of cop peg is less than aluminum hence its resistance.

Question 12.

A parallel plate capacitor consists of two metallic plates separated by a small distance with a dielectric in between.

a. A parallel plate air capacitor has charge densities +a and <r on the plates. Write the expression for the electric field between the plates. What happens to the field if the separation between the plates is doubled ? 1

b. You are given two capacitors. They can be used individually, in series or in parallel in a circuit. Let the four possible values of capacitances be 3 μ F, 4 μ F, 12 μ F, and 16 μ F. If so what are the values of individual capacitances given to you? 2

Answer:

a. , No change in an electric field if plate are separated.

b. We get maximum capacitance in parallel

Similarly, we get minimum capacitance in series

Substituting (1) in (2)

We know

from eqn (1) c_{1} + c_{2} = 16

∴ from eq (1) and (4)

c_{2} = 12μF

Substituting this values in eq(1), we get 16 = 12 + c_{2}

c_{2} = 4pF

Hence answer is 12 μ F and 4 μ F.

Question 13.

A capacitor is a system of two conductors separated by an insulator.

a. Write down the relation for the capacity of a parallel plate capacitor. 1

b. The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5mm. What would be its capacity? 2

Answer:

**Long Answer Type Questions (Score 4)**

Question 1.

In symmetric charge configurations, the electric field can be easily calculated using Gauss’s law. According to Gauss’s law,

a. Obtain an expression for electric field at a point P due to a thin shell of radius R, when the point is at a distance r from the centre of the shell. 2

b. A sphere of radius ‘a‘ is made of insulating material and has a charge distributed uniformly throughout its volume. Let the charge density be p. Find the field due to the charge for r < a. 2

Answer:

a. inside the sphere

b. Choose our Gaussian surface to be a sphere of radius r < a The flux through the Gaussian surface / is φ = E(4π^{2})

The charge distribution is uniform. Therefore, the charge enclosed in the Gaussian sphere of radius r is

This becomes

The magnitude of the electric field in region

Question 2.

One Can determine the direction of the electric field around a stationary charge with the help of electric field lines.

a. What do you understand by the term ‘electric flux'? Give its SI unit. 1

b. Consider a spherical shell of radius ‘R’ is uniformly charged with charge ‘+q’. By using Gauss's theorem, find the electric field intensity at a point ‘p’.

i. Outside this spherical shell

ii. inside this Spherical shell 3

Answer:

a. Electric flux is the total number of electric lines of force passing through a given area. Unit = Nm^{2}/C.

b. Electric flux

i. At a point outside the shell, φ = Eπ^{2} ............ (1)

charge enclosed by the surface = q using Gauss’s theorem

Electric flux ..................(2)

From (1) & (2);

ii. When the point is inside the shell, E= 0 (Since it does not enclose any charge)

Question 3.

For a parallel plate capacitor with each plate of area ‘A‘ separated by distance ‘d‘ in the air, its capacitance is given by

a. Represent the charge ‘q‘ given to a capacitor of capacitance ‘C’ with potential difference ‘V‘ in a graph. What will be the shape of the curve? 1

b. Using the above expression, show that the energy density of a parallel plate capacitor is 1/2 e_{0} E^{2} where ’E’ is the electric field between parallel plates. 3

Answer:

a. The shape of the graph is a straight line.

b. Let a capacitor be charged to a charge Q’. If a small additional charge dQ’ given, the work done will be,

eqn (2) becomes,

Energy per unit volume

Question 4.

Fill in the blanks

Answer:

a.

b. electrovolt

c.

d. Electric susceptibility

Question 5.

VandeGraff Generator is the above device where can develop very high electrical potential.

a. Explain the working principle of the above device 3

b. Is it possible to obtain (-) as well as (+) potential using the above device? Justify. 1

Answer:

a. The Van de Graaff generator works on the following two principles.

1. Discharging action sharp points i.e., electric discharge takes place in air or gases readily at pointed conductors.

2. If the charged conductor is brought into internal contact with a hollow conductor, all of its charges transfer to the surface of the hollow conductor no matter how high the potential of the latter may be.

**Working:**

The high electric field at the pointed ends of comb C, air near them. The +ve charges in the air are repelled and got deposited on the belt through a corona discharge. The charges are carried up to C_{2}. A similar corona discharge takes place at C_{2} and the charges are finally transferred to the shell M. The charges spread over uniformly on the outer surface of M raising its potential to few million volts.

b. Yes. When we supply a positive charge, we will obtain positive potential and vice versa.

Question 6.

The figure shows a parallel plate air capacitor of place area of 100 cm^{2} and separation 4mm. A potential difference of 200V is established between its plates by a battery.

a. Calculate the capacitance and charge on the capacitor 2

b. After disconnecting the battery, the space between the plates is filled by ebonite (k = 2.6). The calculate the capacitance and charge on a capacitor. 2

Answer:

Question 7.

The ability to store electric charge measures the capacitance of a conductor.

a. The SI unit of capacitance is.

b. Show that the energy density of a parallel plate capacitor is

Answer:

a. Farad (F)

b. The energy stored in a capacitor

Energy per unit volume

Question 8.

Explain what would happen to the potential difference between the plates if, in a capacitor of capacitance 1.771 × 10^{11} F, 3 mm thick mica sheet (of dielectric constant = 6) is inserted between the plates:

a. When a supply voltage of 100V is connected. 2

b. When the supply voltage is disconnected. 2

Answer:

a. The dielectric constant of mica sheet = 6

Initial capacitance C = 1.771 × 10^{11} F

Supply voltage V = 100 V

So the potential difference between the plates remains 100 V

But capacitance

C = 6 × 1.771 × 10^{11 }= 10.626 × 10^{11} F The charge on the plates Q = CV = 1.771 × 10^{11} × 100 = 1.771 × 10^{-9} C

b. When the supply voltage is removed. There will be no effect on the charge on the plate.

ie., Q = 1.771 × 10^{-9} C The capacitance

C = 10.626 × 10^{-11} F

Now the potential across the plate is

**Edumate Questions And Answers**

Question 9.

When a charged particle moves in an electric field, work is done on the particle.

a. Pick out a vector quantity for the following

i. electric potential

ii. electric potential gradient

iii. electric potential energy

v. electric flux

b. 5 J of work is done in moving a positive charge of 0.5C between two points. What is the potential difference between the points?

(i) 2.5 V

(ii) 10 V

(iii) 0.1 V

(iv) 5.5 V

c. Three-point electric charges q_{1}= 6 μ C, q_{2} = 4 μ C and q_{3} = -8 μ C are placed on the circumstances of a circle of radius I m as shown in the figure. What is the value of the charge q4 placed on the circle if the potential at the centre of the circle is zero?

Answer:

a. Electric potential gradient

b. 10 V

c. a radius of the circle r = 1m

By the given condition

Question 10.

An equipotential surface is a surface on which the electric potential is the same at every point

a. “Electric field lines are always parallel to Equipotential surfaces”. Cor. direct the statement if there is any mistake.

b. Draw the equipotentials for a single positive point charge.

c. A point charge +q is placed at the centre of a sphere of radius R. Another point charge +q is taken from a point A to another diametrically opposite point B on the surface of the sphere. Calculate the work done for this.

Answer:

a. Electric field lines are always perpendicular to Equipotential surfaces.

b. A surface of the sphere is Equipotential in nature. So the potential difference between any points is zero. Since work done is the product of charge moving and the potential difference between the points, work done is equal to zero.

Question 11.

The capacitance of a capacitor depends on the size and shape of the conductors and on the dielectric material between them.

a. Draw the symbol of a variable capacitor.

b. The plates of a parallel plate capacitor are connected to an ideal voltmeter. What will happen to the reading of the voltmeter if the plate of the capacitor is brought closer to each other in an insulating medium?

c. the plates of a parallel plate capacitor in a vacuum are 5mm apart and 1.5m2 in an area. A potential difference of 10 kV applied across the capacitor. Calculate

i. The capacitance

ii. The charge on each plate

iii. The magnitude of the intensity of the electric field between the plates.

Answer:

b. Capacitance C =

Potential difference

In an insulating medium charge remains the same, voltage is directly proportional to plate separation.

Therefore, voltmeter reading decreases

Question 12.

Capacitors can be combined to obtain any desired capacitance in an application.

a. You are given two capacitors of capacitance 20 μ F each. Draw a diagram to show how you will connect these capacitors to get 40 μ F capacitance.

b. Two capacitors C_{1} and C_{2} are connected in series. This combination is connected in parallel with a third capacitor C_{3}

i. Draw a diagram of the above combination

ii. lf C_{1} = 4 μ F, C_{2} = 6 μ F and C_{3} = 2.4 μ F and a potential of 100 V is applied across the combination calculate the charge stored in each capacitor.

Answer:

a.

c. Effective capacitance of C_{1} and C_{2} Potential difference across C_{12} = 100V Charge stored on C_{12}, q_{12} = C_{12} × 100

q_{12} = 2.4 × 10^{-6} × 100

q_{12} = 2.4 × 10^{-4 }C

Charge on C_{1}

q_{1} = 2.4 × 10^{-4 }C

Charge on C_{2}

q_{2} = 2.4 × 10^{-4} C

Potential differences across C_{3} = 100V

Charge on C_{3}

q_{3} = C_{3} × 100

q_{3} = 2.4 × 10^{-6} × 100

q_{3 }= 2.4 × 10^{-4 }C

Question 13.

Many of the applications of capacitors depend on their ability to store energy.

a. In a charged capacitor energy is stored in the

(i) Positively charged plate

(ii) Negatively charged plate

(iii) Electric field between the plate None of these

b. Draw a graph showing the variation of charge stored in a capacitor with its potential. How will you calculate the energy stored in the capacitor using the above graph?

c. An electric flash lamp has 20 capacitors each of capacitance 5 μ F connected in parallel. The lamp is operated at 100V. If the energy stored in the combination is completely radiated out in a single flash, how much energy will be radiated in a flash?

Answer:

a. An electric field between the plates.

b. An area under the straight line graph gives the energy stored in the capacitor.

c. Effective capacitance C = 20 × 5 μ F = 100 μ F

Energy stored

**Ncert Questions And Answers**

Question 1.

Capacitors are devices used to store charges.

Capacitors in series

a. What happens to the capacitance of the capacitor if the medium in between the plates is filled with a dielectric of relative permittivity?

b. If three capacitances C_{1} C_{2} and C_{3} are connected in series, what is the effective capacitance of the combination?

c. It is not practicable to construct a capacitor with capacitance 1 Farad and distance between the plates 1 m Why?

Answer:

a. Capacitance increases

b.

c. The capacitance of a spherical conductor

∴ The radius of a spherical conductor

r = 9 × 10^{6} km, which is not possible.

Question 2.

Two charges 5 × 10^{-8}C and -3 × 10^{-8 }C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Answer:

There are two charges, q_{1} = 5 × 10^{-8 }C, q_{2} = 3 × 10^{-8}C

Distance between the two charges, d = 16 cm = 0.16m

Consider a point P on the line joining the two charges, as shown in the given figure.

r = Distance of point P from charge q_{1}

Let the electric potential (V) at point P be zero. A potential at point P is the sum of potentials caused by charges g_{1} and g_{2} respectively.

Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges.

Suppose point P is outside the system of two charges at a distance S from the positive charge, where potential is zero, as shown in the following figure.

Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.

Question 3.

Two charges 2 μ C and -2 μ C are placed at points A and B is 6 cm apart.

a. Identify an equipotential surface of the system.

b. What is the direction of the electric field at every point on this surface?

Answer:

a. The situation is represented in the given figure.

An equipotential surface is a plane on which total potential is zero everywhere. This plane is normal to line AB. The plane is located at the midpoint of line AB because the magnitude of charges is the same, b. The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.

Question 4.

A regular hexagon of side 10 cm has a charge 5 μ C at each of its vertices. Calculate the potential at the centre of the hexagon.

Answer:

Question 5.

A spherical conductor of radius 12 cm has a charge of 1.6 × 10^{-7 }C distributed uniformly on its surface. What is the electric field:

a. Inside the sphere?

b. Just outside the sphere?

c. At a point 18 cm from the centre of the sphere?

Answer:

a. Radius of the spherical conductor, r = 12 cm = 0.12 m Charge is uniformly distributed over the

conductor, q = 1.6 × 10^{-7 }C Electric field inside a spherical conductor is zero. This is because if there is a field inside the conductor, then charges will move to neutralize it.

b. Electric field E just outside the conductor is given by the relation,

c. The electric field at a point 18 m from the centre of the sphere = E_{1}

Distance of the point from the centre, d= 18 cm = 0.18 m

The electric field at a point 18 cm from the centre of the sphere is 4.4 × 10^{4} N/C

Question 6.

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (lpF = 10^{-12} F). 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:

A capacitance between the parallel plates of the capacitor, C = 8 pF

Initially, the distance between the parallel plates was d and it was filled with air. A dielectric constant of air, k = 1,

Capacitance, C', is given by the formula,

where A = Area of each plate, ∈0 = Permittivity of free space If a distance between the plates is reduced to half then new distance, d = a/2 Dielectric constant of the substance filled in between the plates, k = 6 Hence, a capacitance of the capacitor becomes

Taking ratios of equations (i) and (ii), we j obtain C°= 2 × 6C = 12 C= 12 × 8 = 96 pF

Therefore, the capacitance between the plates is 96 pF.

Question 7.

Three capacitors each of capacitance 9 pF 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:

a. A capacitance of each of the three capacitors, C = 9 pF

Equivalent capacitance ( c ’) of the combination of the capacitors is given by the relation,

Therefore, the total capacitance of the combination is 3 pF.

Supply voltage, V= 120 V. Potential difference (V) across each capacitor is equal to one-third of the supply voltage.

Therefore, the potential difference across each capacitor is 40 V.

Question 8.

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

Answer:

The capacitance of the capacitor, C = 600 pF Potential difference, V= 200 V Electrostatic energy stored in the capacitor is given by,

If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (c') of the combination is given by,

New electrostatic energy can be calculated as

Loss in electrostatic energy = E - E'

= 1.2 × 10^{-5 }-6.0 × 10^{-5}

= 0.6 × 10^{-5} = 6 × 10^{-6
}Therefore, the electrostatic energy lost in the process is 6 × 10^{-6} J.

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