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Class 12 physics chapter 1 NCERT solutions: Electric Charges and Fields PDF

Class 12 physics chapter 1 NCERT solutions: Electric Charges and Fields PDF: Here get all the NCERT solutions for all questions on the book. Also get explanation video after each question. If you have any doubt then please don’t hesitate to ask. Download the PDF for the whole chapter so that you can see the answers when you are offline to. Here is the whole Exercise of NCERT solutions.

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Class 12 physics chapter 1 NCERT solutions: EXERCISES

Question: 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, r = 30 cm = 0.3 m

Electrostatic force between the spheres is given by the relation

Where, ∈0 = Permittivity of free space

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 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:

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,

Where, ∈0 = Permittivity of free space

The distance between the two spheres is 0.12 m.

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 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 .

 

Where,

G = Gravitational constant

Its unit is N m2 kg−2.

me and mp = Masses of electron and proton. Their unit is kg.

e = Electric charge. Its unit is C.

∈0 = Permittivity of free space.

Its unit is N m2 C−2.

Hence, the given ratio is dimensionless.

e = 1.6 × 10−19 C

G = 6.67 × 10−11 N m2 kg-2

me= 9.1 × 10−31 kg

mp = 1.66 × 10−27 kg

Hence, the numerical value of the given ratio is

 

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

 

Question 4:

Explain the meaning of the statement ‘electric charge of a body is quantised’. Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Answer:

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.

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 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 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 nature 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 6:

Four point charges qA = 2 μC, qB = −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 center of the square?

Answer:

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

Where,

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

(Diagonals) AC = BD =      10 by root 2     cm

 

AO = OC = DO = OB = 5 by root 2  cm

 

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

 

Force of repulsion between charges placed at corner A and center O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and center O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and center O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and center 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 center O is zero.

Question 7:

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

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

 

Answer:

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.

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 8:

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

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

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:

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

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,

E=   along OB

= Permittivity of free space

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

E2 =along OB

 

= 5.4 × 106 N/C along OB

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

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.

 

Question 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.

At A, amount of charge, qA = 2.5 × 10−7C 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 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

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

Class 12 physics chapter 1 NCERT solutions VIDEOS.

Question 11:

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

Estimate the number of electrons transferred (from which to which?) Is there a transfer of mass from wool to polythene?

Answer:

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

n can be calculated using the relation,

q = ne

= 1.87 × 1012

 

Therefore, the number of electrons transferred from wool to polythene is 1.87 × 1012. 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 12:

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.

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

Answer:

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,

Where

∈0 = Free space permittivity

= 1.52 × 10−2 N

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

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.

Force of repulsion between the two spheres,

= 16 × 1.52 × 10−2

= 0.243 N

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

Class 12 physics chapter 1 NCERT solutions VIDEOS.

Question 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, amount of charge from A will

transfer to sphere C. Hence, charge on each of the spheres, A and C, is

When sphere C with charge

is brought in contact with sphere B with charge q, total Charges on the system will divide into two equal halves given as,

Each sphere will each half. Hence, charge on each of the spheres, C and B, is

Force of repulsion between sphere A having charge and sphere B having charge =

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

Question 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?

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 15:

Consider a uniform electric field E = 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:

Electric field intensity, E = 3 × 103 î N/C

Magnitude of electric field intensity,   E = 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 =

= 3 × 103 × 0.01 × cos0°

= 30 N m2/C

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

Flux, Φ =

= 3 × 103 × 0.01 × cos60°

= 15 N m2/C

Question 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.

 

Class 12 physics chapter 1 NCERT solutions Videos click here.

 

 

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