NCERT Solutions for Class 12 Physics Chapter 1 – Electric Charges And Fields

Page No 46:

Question 1.1:

What is the force between two small charged spheres having charges of 2 × 10−7 C and 3 × 10−7 C placed 30 cm apart in air?

Answer:

Repulsive force of magnitude 6 × 10−3 N

Charge on the first sphere, q1 = 2 × 10−7 C

Charge on the second sphere, q2 = 3 × 10−7 C

Distance between the spheres, = 30 cm = 0.3 m

Electrostatic force between the spheres is given by the relation,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 2ab5d65b

Where, ∈0 = Permittivity of free space

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 2b4c6d34

Hence, force between the two small charged spheres is 6 × 10−3 N. The charges are of same nature. Hence, force between them will be repulsive.

Question 1.2:

The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Answer:

(a) Electrostatic force on the first sphere, F = 0.2 N

Charge on this sphere, q1 = 0.4 μC = 0.4 × 10−6 C

Charge on the second sphere, q2 = − 0.8 μC = − 0.8 × 10−6 C

Electrostatic force between the spheres is given by the relation,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 2ab5d65bNS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 6c0ed7d7

Where, ∈0 = Permittivity of free space

r2 = q1q24π∈0F    = 9×109×0.4×10-6×0.8×10-60.2     = 144×10-4r   = 144×10-4 = 0.12 m

The distance between the two spheres is 0.12 m.

(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.

Question 1.3:

Check that the ratio ke2/G memp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Answer:

The given ratio is NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 41d5f90c.

Where,

G = Gravitational constant

Its unit is N mkg−2.

me and mp = Masses of electron and proton.

Their unit is kg.

e = Electric charge.

Its unit is C.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 5812a67e

0 = Permittivity of free space

Its unit is N mC−2.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m59350536

Hence, the given ratio is dimensionless.

= 1.6 × 10−19 C

G = 6.67 × 10−11 N mkg-2

me= 9.1 × 10−31 kg

mp = 1.66 × 10−27 kg

Hence, the numerical value of the given ratio is

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 6097e97a

This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

Question 1.4:

(a) Explain the meaning of the statement ‘electric charge of a body is quantised’.

(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Answer:

(a) Electric charge of a body is quantized. This means that only integral (1, 2, …., n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.

(b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.

Question 1.5:

When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Answer:

Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

Question 1.6:

Four point charges q= 2 μC, q= −5 μC, qC = 2 μC, and qD = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

Answer:

The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m365d3784

Where,

(Sides) AB = BC = CD = AD = 10 cm

(Diagonals) AC = BD = NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m30eeb009cm

AO = OC = DO = OB = NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 287b5bbecm

A charge of amount 1μC is placed at point O.

Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 μC charge at centre O is zero.

Question 1.7:

(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

(b) Explain why two field lines never cross each other at any point?

Answer:

(a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.

(b) If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.

Question 1.8:

Two point charges qA = 3 μC and qB = −3 μC are located 20 cm apart in vacuum.

(a) What is the electric field at the midpoint O of the line AB joining the two charges?

(b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge?

Answer:

(a) The situation is represented in the given figure. O is the mid-point of line AB.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 60d2b61f

Distance between the two charges, AB = 20 cm

∴AO = OB = 10 cm

Net electric field at point O = E

Electric field at point O caused by +3μC charge,

E1 = NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 15b6e68c along OB

Where,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 149b60d9= Permittivity of free space

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 793ab607

Magnitude of electric field at point O caused by −3μC charge,

E2 = NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 4829d0c5 = NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 621dee13 along OB

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m4cbbecde

= 5.4 × 106 N/C along OB

Therefore, the electric field at mid-point O is 5.4 × 106 N C−1 along OB.

(b) A test charge of amount 1.5 × 10−9 C is placed at mid-point O.

q = 1.5 × 10−9 C

Force experienced by the test charge = F

F = qE

= 1.5 × 10−9 × 5.4 × 106

= 8.1 × 10−3 N

The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.

Therefore, the force experienced by the test charge is 8.1 × 10−3 N along OA.

Question 1.9:

A system has two charges qA = 2.5 × 10−7 C and qB = −2.5 × 10−7 C located at points A: (0, 0, − 15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moment of the system?

Answer:

Both the charges can be located in a coordinate frame of reference as shown in the given figure.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 468318da

At A, amount of charge, qA = 2.5 × 107C

At B, amount of charge, qB = −2.5 × 10−7 C

Total charge of the system,

q = qA + qB

= 2.5 × 107 C − 2.5 × 10−7 C

= 0

Distance between two charges at points A and B,

d = 15 + 15 = 30 cm = 0.3 m

Electric dipole moment of the system is given by,

p = qA × d = qB × d

= 2.5 × 10−7 × 0.3

= 7.5 × 10−8 C m along positive z-axis

Therefore, the electric dipole moment of the system is 7.5 × 10−8 C m along positive z−axis.

Question 1.10:

An electric dipole with dipole moment 4 × 10−9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104 N C−1. Calculate the magnitude of the torque acting on the dipole.

Answer:

Electric dipole moment, p = 4 × 10−9 C m

Angle made by p with a uniform electric field, θ = 30°

Electric field, E = 5 × 104 N C−1

Torque acting on the dipole is given by the relation,

τ = pE sinθ

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m6822f3ba

Therefore, the magnitude of the torque acting on the dipole is 10−4 N m.

Question 1.11:

A polythene piece rubbed with wool is found to have a negative charge of 3 × 10−7 C.

(a) Estimate the number of electrons transferred (from which to which?)

(b) Is there a transfer of mass from wool to polythene?

Answer:

(a) When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged.

Amount of charge on the polythene piece, q = −3 × 10−7 C

Amount of charge on an electron, e = −1.6 × 10−19 C

Number of electrons transferred from wool to polythene = n

can be calculated using the relation,

q = ne

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 20267e16

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m6985479a

= 1.87 × 1012

Therefore, the number of electrons transferred from wool to polythene is 1.87 × 1012.

(b) Yes.

There is a transfer of mass taking place. This is because an electron has mass,

me = 9.1 × 10−3 kg

Total mass transferred to polythene from wool,

m = me × n

= 9.1 × 10−31 × 1.85 × 1012

= 1.706 × 10−18 kg

Hence, a negligible amount of mass is transferred from wool to polythene.

Question 1.12:

(a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10−7 C? The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Answer:

(a) Charge on sphere A, qA = Charge on sphere B, qB = 6.5 × 10−7 C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between the two spheres,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m646a4f7

Where,

0 = Free space permittivity

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html md1bb365= 9 × 109 N m2 C−2

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 4a701455

= 1.52 × 10−2 N

Therefore, the force between the two spheres is 1.52 × 10−2 N.

(b) After doubling the charge, charge on sphere A, qA = Charge on sphere B, qB = 2 × 6.5 × 10−7 C = 1.3 × 10−6 C

The distance between the spheres is halved.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 6983d7a0

Force of repulsion between the two spheres,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m646a4f7

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 1785d2b0

= 16 × 1.52 × 10−2

= 0.243 N

Therefore, the force between the two spheres is 0.243 N.

Page No 47:

Question 1.13:

Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Answer:

Distance between the spheres, A and B, r = 0.5 m

Initially, the charge on each sphere, q = 6.5 × 10−7 C

When sphere A is touched with an uncharged sphere C, NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 352eec5c amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, isNS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 352eec5c.

When sphere C with charge NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 352eec5c is brought in contact with sphere B with charge q, total charges on the system will divide into two equal halves given as,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m5839ca28

Each sphere will share each half. Hence, charge on each of the spheres, C and B, isNS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m570ffa7e.

Force of repulsion between sphere A having charge NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 352eec5c and sphere B having charge NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m570ffa7e =NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 2fc0c0a3

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html mf1f083

Therefore, the force of attraction between the two spheres is 5.703 × 10−3 N.

Question 1.14:

Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 2c6b2781

Answer:

Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.

The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

Question 1.15:

Consider a uniform electric field = 3 × 103 îN/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

Answer:

(a) Electric field intensity, NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 2d5b349 = 3 × 10î N/C

Magnitude of electric field intensity, NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 132ee3c1= 3 × 103 N/C

Side of the square, s = 10 cm = 0.1 m

Area of the square, A = s2 = 0.01 m2

The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, θ = 0°

Flux (Φ) through the plane is given by the relation,

Φ = NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m58ea659b

= 3 × 103 × 0.01 × cos0°

= 30 N m2/C

(b) Plane makes an angle of 60° with the x-axis. Hence, θ = 60°

Flux, Φ = NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m58ea659b

= 3 × 103 × 0.01 × cos60°

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 654e1886 = 15 N m2/C

Question 1.16:

What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Answer:

All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.

Question 1.17:

Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 103 N m2/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

Answer:

(a) Net outward flux through the surface of the box, Φ = 8.0 × 103 N m2/C

For a body containing net charge q, flux is given by the relation,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 1cab0b31

0 = Permittivity of free space

= 8.854 × 10−12 N−1Cm−2

q = ∈0Φ

= 8.854 × 10−12 × 8.0 × 103

= 7.08 × 10−8

= 0.07 μC

Therefore, the net charge inside the box is 0.07 μC.

(b) No

Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.

Question 1.18:

A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m5f4bf5d6

Answer:

The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem for a cube, total electric flux is through all its six faces.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 13008a7a

Hence, electric flux through one face of the cube i.e., through the square, NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m34d1b516

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m28835193

Where,

0 = Permittivity of free space

= 8.854 × 10−12 N−1Cm−2

q = 10 μC = 10 × 10−6 C

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html mc736652

= 1.88 × 105 N m2 C−1

Therefore, electric flux through the square is 1.88 × 105 N m2 C−1.

Page No 48:

Question 1.19:

A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Answer:

Net electric flux (ΦNet) through the cubic surface is given by,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html e8b2bee

Where,

0 = Permittivity of free space

= 8.854 × 10−12 N−1Cm−2

q = Net charge contained inside the cube = 2.0 μC = 2 × 10−6 C

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 4a6125d

= 2.26 × 105 N m2 C−1

The net electric flux through the surface is 2.26 ×105 N m2C−1.

Question 1.20:

A point charge causes an electric flux of −1.0 × 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

Answer:

(a) Electric flux, Φ = −1.0 × 103 N m2/C

Radius of the Gaussian surface,

r = 10.0 cm

Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., −103 N m2/C.

(b) Electric flux is given by the relation,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 1cab0b31

Where,

q = Net charge enclosed by the spherical surface

0 = Permittivity of free space = 8.854 × 10−12 N−1Cm−2

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 5cae3633

= −1.0 × 103 × 8.854 × 10−12

= −8.854 × 10−9 C

= −8.854 nC

Therefore, the value of the point charge is −8.854 nC.

Question 1.21:

A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10N/C and points radially inward, what is the net charge on the sphere?

Answer:

Electric field intensity (E) at a distance (d) from the centre of a sphere containing net charge q is given by the relation,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 77b12b42

Where,

q = Net charge = 1.5 × 103 N/C

= Distance from the centre = 20 cm = 0.2 m

0 = Permittivity of free space

And,NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html md1bb365 = 9 × 109 N m2 C−2

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 5afed959

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m5ee0863f

= 6.67 × 109 C

= 6.67 nC

Therefore, the net charge on the sphere is 6.67 nC.

Question 1.22:

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

Answer:

(a) Diameter of the sphere, d = 2.4 m

Radius of the sphere, r = 1.2 m

Surface charge density, NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m5dfc57c4= 80.0 μC/m2 = 80 × 10−6 C/m2

Total charge on the surface of the sphere,

Q = Charge density × Surface area

=NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m3d0eadd7

= 80 × 10−6 × 4 × 3.14 × (1.2)2

= 1.447 × 10−3 C

Therefore, the charge on the sphere is 1.447 × 10−3 C.

(b) Total electric flux (NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m41ac72c) leaving out the surface of a sphere containing net charge is given by the relation,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 30aecee1

Where,

0 = Permittivity of free space

= 8.854 × 10−12 N−1Cm−2

= 1.447 × 10−3 C

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 73d9b193

= 1.63 × 108 N C−1 m2

Therefore, the total electric flux leaving the surface of the sphere is 1.63 × 108 N C−1 m2.

Question 1.23:

An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.

Answer:

Electric field produced by the infinite line charges at a distance d having linear charge density λ is given by the relation,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 413a30c4

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m389953f4

Where,

d = 2 cm = 0.02 m

E = 9 × 104 N/C

0 = Permittivity of free space

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html md1bb365 = 9 × 109 N m2 C−2

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m34b2ebed

= 10 μC/m

Therefore, the linear charge density is 10 μC/m.

Question 1.24:

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10−22 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

Answer:

The situation is represented in the following figure.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m4ccabe7

A and B are two parallel plates close to each other. Outer region of plate A is labelled as I, outer region of plate B is labelled as III, and the region between the plates, A and B, is labelled as II.

Charge density of plate A, σ = 17.0 × 10−22 C/m2

Charge density of plate B, σ = −17.0 × 10−22 C/m2

In the regions, I and III, electric field E is zero. This is because charge is not enclosed by the respective plates.

Electric field E in region II is given by the relation,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 56715273

Where,

0 = Permittivity of free space = 8.854 × 10−12 N−1Cm−2

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 5cc70304

= 1.92 × 10−10 N/C

Therefore, electric field between the plates is 1.92 × 10−10 N/C.

Question 1.25:

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C−1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm−3. Estimate the radius of the drop. (= 9.81 m s−2= 1.60 × 10−19 C).

Answer:

Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 × 104 N C−1

Density of oil, ρ = 1.26 gm/cm3 = 1.26 × 103 kg/m3

Acceleration due to gravity, g = 9.81 m s−2

Charge on an electron, e = 1.6 × 10−19 C

Radius of the oil drop = r

Force (F) due to electric field E is equal to the weight of the oil drop (W)

F = W

Eq mg

Ene NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m5e59792a

Where,

q = Net charge on the oil drop = ne

= Mass of the oil drop

= Volume of the oil drop × Density of oil

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 2861d474

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html mcf8da84

= 9.82 × 10−4 mm

Therefore, the radius of the oil drop is 9.82 × 10−4 mm.

Question 1.26:

Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

(a)

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 357bb3da

(b)

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 5d9aa60

(c)

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 674b3e39

(d)

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 773383f

(e)

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html be1906a

Answer:

(a) The field lines showed in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.

(b) The field lines showed in (b) do not represent electrostatic field lines because the field lines cannot emerge from a negative charge and cannot terminate at a positive charge.

(c) The field lines showed in (c) represent electrostatic field lines. This is because the field lines emerge from the positive charges and repel each other.

(d) The field lines showed in (d) do not represent electrostatic field lines because the field lines should not intersect each other.

(e) The field lines showed in (e) do not represent electrostatic field lines because closed loops are not formed in the area between the field lines.

Page No 49:

Question 1.27:

In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 105 NC−1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10−7 Cm in the negative z-direction?

Answer:

Dipole moment of the system, p = q × dl = −10−7 C m

Rate of increase of electric field per unit length,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 56bf22cf

Force (F) experienced by the system is given by the relation,

F = qE

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m4c5f56f6

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 4d2774f8

= −10−7 × 10−5

= −10−2 N

The force is −10−2 N in the negative z-direction i.e., opposite to the direction of electric field. Hence, the angle between electric field and dipole moment is 180°.

Torque (τ) is given by the relation,

τ = pE sin180°

= 0

Therefore, the torque experienced by the system is zero.

Question 1.28:

(a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is [Fig. 1.36(b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m455a294c

Answer:

(a) Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity. The electric field intensity E inside the charged conductor is zero.

Let q is the charge inside the conductor and NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 149b60d9is the permittivity of free space.

According to Gauss’s law,

Flux, NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m378d9655

Here, = 0

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m1e5b959c

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 74e530e

Therefore, charge inside the conductor is zero.

The entire charge Q appears on the outer surface of the conductor.

(b) The outer surface of conductor A has a charge of amount Q. Another conductor B having charge +q is kept inside conductor A and it is insulated from A. Hence, a charge of amount −will be induced in the inner surface of conductor A and +q is induced on the outer surface of conductor A. Therefore, total charge on the outer surface of conductor A is Q q.

(c) A sensitive instrument can be shielded from the strong electrostatic field in its environment by enclosing it fully inside a metallic surface. A closed metallic body acts as an electrostatic shield.

Question 1.29:

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 87ddd34NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 21845f56, where NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 21845f56is the unit vector in the outward normal direction, and NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m5dfc57c4 is the surface charge density near the hole.

Answer:

Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.

Let E is the electric field just outside the conductor, q is the electric charge, NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m5dfc57c4is the charge density, and NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 149b60d9is the permittivity of free space.

Charge NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m2ab93157

According to Gauss’s law,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m7ab421ac

Therefore, the electric field just outside the conductor isNS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 68f895e4. This field is a superposition of field due to the cavity NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 4414483dand the field due to the rest of the charged conductorNS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 4414483d. These fields are equal and opposite inside the conductor, and equal in magnitude and direction outside the conductor.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 28c801ac

Therefore, the field due to the rest of the conductor isNS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 68f895e4.

Hence, proved.

Question 1.30:

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Answer:

Take a long thin wire XY (as shown in the figure) of uniform linear charge densityNS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m11cc021f.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m598b777

Consider a point A at a perpendicular distance l from the mid-point O of the wire, as shown in the following figure.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m4f3b4c81

Let E be the electric field at point A due to the wire, XY.

Consider a small length element dx on the wire section with OZ = x

Let q be the charge on this piece.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m663934a4

Electric field due to the piece,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m3c100072

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 619d290a

The electric field is resolved into two rectangular components. NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m4328ddb0 is the perpendicular component and NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m4d6a9b96 is the parallel component.

When the whole wire is considered, the component NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m4d6a9b96 is cancelled.

Only the perpendicular component NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m4328ddb0 affects point A.

Hence, effective electric field at point A due to the element dx is dE1.

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 3dc44703

On differentiating equation (2), we obtain

dxdθ = lsec2θdx =  lsec2θ dθ

From equation (2),

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m7a1f7e58

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m52087de4

Putting equations (3) and (4) in equation (1), we obtain

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m2f223adf

The wire is so long that NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 69d7c482 tends from NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m2f290e5cto NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m6f21b139.

By integrating equation (5), we obtain the value of field E1 as,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 2e68ce1b

Therefore, the electric field due to long wire isNS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m1d802896.

Question 1.31:

It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge (+2/3) e, and the ‘down’ quark (denoted by d) of charge (−1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

Answer:

A proton has three quarks. Let there be n up quarks in a proton, each having a charge of NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html mc6497ff.

Charge due to up quarksNS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m4ca7d9a3

Number of down quarks in a proton = 3 − n

Each down quark has a charge of NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 49d3924f.

Charge due to (3 − n) down quarks NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 56ab3580

Total charge on a proton = + e

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m39bd23c3

Number of up quarks in a proton, n = 2

Number of down quarks in a proton = 3 − n = 3 − 2 = 1

Therefore, a proton can be represented as ‘uud’.

A neutron also has three quarks. Let there be n up quarks in a neutron, each having a charge of NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 1b8cec93.

Charge on a neutron due to n up quarks NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 1f1774fc

Number of down quarks is 3 − n,each having a charge of NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 4eee044c.

Charge on a neutron due to NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 47876daadown quarks = NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m79eb00d3

Total charge on a neutron = 0

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 745b3a

Number of up quarks in a neutron, n = 1

Number of down quarks in a neutron = 3 − n = 2

Therefore, a neutron can be represented as ‘udd’.

Page No 50:

Question 1.32:

(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.

(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

Answer:

(a) Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.

(b) Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.

Question 1.33:

A particle of mass and charge (−q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig. 1.33). The length of plate is and an uniform electric field is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL2/ (2mNS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 1ba67720).

Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.

Answer:

Charge on a particle of mass m = − q

Velocity of the particle = vx

Length of the plates = L

Magnitude of the uniform electric field between the plates = E

Mechanical force, F = Mass (m) × Acceleration (a)

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 44644716

Therefore, acceleration, NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html m1730b78

Time taken by the particle to cross the field of length is given by,

tNS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 3f9d7ad5

In the vertical direction, initial velocity, u = 0

According to the third equation of motion, vertical deflection of the particle can be obtained as,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html mcc380a2

Hence, vertical deflection of the particle at the far edge of the plate is

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 76adf2cc. This is similar to the motion of horizontal projectiles under gravity.

Question 1.34:

Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx= 2.0 × 106 m s−1. If between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate? (| | =1.6 × 10−19 C, m= 9.1 × 10−31 kg.)

Answer:

Velocity of the particle, vx = 2.0 × 106 m/s

Separation of the two plates, d = 0.5 cm = 0.005 m

Electric field between the two plates, E = 9.1 × 102 N/C

Charge on an electron, q = 1.6 × 10−19 C

Mass of an electron, m= 9.1 × 10−31 kg

Let the electron strike the upper plate at the end of plate L, when deflection is s.

Therefore,

NS 21 10 08 Sravana 12 Physics 1 34 NRJ SS html 13ff71f6

Therefore, the electron will strike the upper plate after traveling 1.6 cm.


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