NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 1: Electric Charges and Fields

Question1.1: What is the force between two small charged spheres having charges of 2 × 10–7 C & 3 × 10–7 C placed 30 cm apart in air?

Solution1.1:

Given:

Q1= 2 × 10-7 C

Q2 = 3 × 10-7 C

= 30 × 10-2 m

We know,

= (Q1 Q2)/(4 π εo r2 )

Substituting the given values, we get

F = 6 × 10-3 N.

Question1.2: 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?

Solution1.2:

(a) From the formula

F = (Q1 Q2)/(4πεo r2 )

Substituting the values, we get

r = 0.12 m = 12 cm.

(b) Force on second sphere due to first

F21 = (Q1 Q2)/(4πεo r2 )

This comes to be 0.2 N which means force on second sphere due to first. It confirms Newton’s third law.

Question 1.3: Check that the ratio {ke2} /{G me mp} is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Solution 1.3:

We know very well

k = 9 × 109 Nm2 C‒2

e = 1.6 × 10‒19 C

G = 6.6 × 10‒11 Nm2 kg‒2

me = 9.11 × 10‒31 kg

mp = 1.6 × 10‒27 kg

Putting these values in{ke2} /{G me mp} its value is 2.29 × 1039.

Ratio = {ke2} /{G me mp} = {[Nm2 C‒2] [C]}/{[Nm2 kg‒2] [kg] [kg]} = 1

Hence, the given ration is dimensionless.

The ratio signifies that electrical forces are immensely stronger than the gravitational forces.

Question1.4 (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?

Solution 1.4:

(a) Quantization is one of the three basic properties of electric charge. It means that every charge is an integral multiple of e, i.e., nen = …-2, -1, 0, 1, 2…The addition of charges, subtraction of charges, being an integer always gives integer result. Thus a charge can always be incremented or decremented in terms of e.

(b) Macroscopic charges have very large number of electrons. The quantization here can be taken as a continuous phenomenon, analogous to closely spaced dots resembling to line from a distance.

Question1.5: 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.

Solution 1.5:

When two bodies are rubbed with each other transfer of charge takes place. One body receives charge and other loses, becoming negatively and positively charged respectively. In the whole process no new charge is created or destroyed. This implies that in an isolated system the total charge is always conserved.

Quesion1.6: 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 centre of the square?

Solution 1.6:

The charges of equal magnitude and same sign are at the corners of same diagonal. So they will exhibit equal and opposite forces at the charge situated at center, cancelling out each other. So, the force is zero Newton.

Question1.7 (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?

Solution 1.7:

(a) The direction of electric field is given by tangent at each point on the curve. At sudden breaks, the field will have more than one direction which is not possible. That’s why electrostatic field line is a continuous curve.

(b) At the crossing point there will be two directions of electric field at that point given by the two tangents. This cannot happen, and so two field lines never cross each other at any point.

Question1.8: Two point charges q= 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?

Solution 1.8:

(a) Given:

qA= 3 μC = 3 × 10‒6 C

qB = ‒3 μC = ‒ 3 × 10‒6 C

= 10 cm (point O is in middle)

We know, electric field is given by formula

E = Q/(4πϵo r2 )

Electric field due to charge A (by using above formula)

|EA| = 2.7 × 106 N / C

Electric field due to charge B

|EB| = 2.7 × 106 N/C opposite to the direction of EA.

NCERT Solutions for CBSE Class 12 Physics, Chapter 1, Question1.8

Since both the fields have the same direction, the electric field at O will be summation of above field.

So, |ENet| = 5.4 × 106 N/C towards B.

(b) Given:

qo = 1.5 × 10–9 C

|ENet| = 5.4 × 106 N/C towards B

By using the formula

F = q E

We have,

F = (qo) (|ENet|) = (1.5 × 10–9 C) × (5.4 × 106 N/C) = 8.1 × 10‒3 towards charge qB  (because direction of electric field is towards B).

Question1.9: 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?

Solution1.9:

Total charge of electric dipole = zero coulomb

Magnitude of Dipole moment,

|p| = (Magnitude of either charge) × (Distance between 2 charges) = q.2a

Given, 2a = 30 cm, q = 2.5 × 10‒7 C

= (2.5 × 10‒7) × (30)

= 7.5 × 10‒8 C-m

Question1.10 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 NC–1. Calculate the magnitude of the torque acting on the dipole.

Solution 1.10:

Dipole moment, |p| = 4 × 10-9 Cm, θ = 30oE = 5 × 104 N/C

We know that,

Torque = p E sinθ

Torque = (4 × 10‒9) (5 ×104) (sin30o)

Torque = 10-4 Nm

Question1.11: A polythene piece rubbed with wool is found to have a ‒ve 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?

Solution 1.11:

Given,

q = -3 × 10-7 C on polythene

Charge on polythene is negative so electrons are transferred from wool to polythene.

(a)q = ne

Since, = 1.6 × 10-19 C

Therefore,n = |q|/= 1.875 × 1012 electrons.

1.875 × 1012 electrons are transferred from wool to polythene.

(b) Yes mass is transferred as electron has a mass of 9.1 × 10‒31 kg.

Total mass transferred

= number of electrons transferred × mass of an electron

= (1.875 × 1012) (9.1 × 10‒31) = 1.7 × 10‒18 kg.

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,

Where,

0 = Free space permittivity

= 9 × 109 N m2 C−2

= 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.

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.

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,  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 share 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 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?

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,  = 3 × 10î N/C

Magnitude of electric field intensity, = 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

(b) 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 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,

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.)

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.

Hence, electric flux through one face of the cube i.e., through the square, 

Where,

0 = Permittivity of free space

= 8.854 × 10−12 N−1Cm−2

q = 10 μC = 10 × 10−6 C

= 1.88 × 105 N m2 C−1

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

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,

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

= 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,

Where,

q = Net charge enclosed by the spherical surface

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

= −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,

Where,

q = Net charge = 1.5 × 103 N/C

= Distance from the centre = 20 cm = 0.2 m

0 = Permittivity of free space

And, = 9 × 109 N m2 C−2

= 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, = 80.0 μC/m2 = 80 × 10−6 C/m2

Total charge on the surface of the sphere,

Q = Charge density × Surface area

=

= 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 () leaving out the surface of a sphere containing net charge is given by the relation,

Where,

0 = Permittivity of free space

= 8.854 × 10−12 N−1Cm−2

= 1.447 × 10−3 C

= 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,

Where,

d = 2 cm = 0.02 m

E = 9 × 104 N/C

0 = Permittivity of free space

 = 9 × 109 N m2 C−2

= 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.

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,

Where,

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

= 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 

Where,

q = Net charge on the oil drop = ne

= Mass of the oil drop

= Volume of the oil drop × Density of oil

= 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)

(b)

(c)

(d)

(e)

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.

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,

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

F = qE

= −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.

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 is the permittivity of free space.

According to Gauss’s law,

Flux, 

Here, = 0

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 , where is the unit vector in the outward normal direction, and  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, is the charge density, and is the permittivity of free space.

Charge 

According to Gauss’s law,

Therefore, the electric field just outside the conductor is. This field is a superposition of field due to the cavity and the field due to the rest of the charged conductor. These fields are equal and opposite inside the conductor, and equal in magnitude and direction outside the conductor.

Therefore, the field due to the rest of the conductor is.

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 density.

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

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.

Electric field due to the piece,

The electric field is resolved into two rectangular components.  is the perpendicular component and  is the parallel component.

When the whole wire is considered, the component  is cancelled.

Only the perpendicular component  affects point A.

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

On differentiating equation (2), we obtain

dxdθ = lsec2θdx =  lsec2θ dθdxdθ = lsec2θdx =  lsec2θ dθ

From equation (2),

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

The wire is so long that  tends from to .

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

Therefore, the electric field due to long wire is.

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 .

Charge due to up quarks

Number of down quarks in a proton = 3 − n

Each down quark has a charge of .

Charge due to (3 − n) down quarks 

Total charge on a proton = + e

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 .

Charge on a neutron due to n up quarks 

Number of down quarks is 3 − n,each having a charge of .

Charge on a neutron due to 

down quarks = 

Total charge on a neutron = 0

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’.

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.

Page No 50:

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/ (2m).

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)

Therefore, acceleration, 

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

t

In the vertical direction, initial velocity, u = 0

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

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

. This is similar to the motion of horizontal projectiles under gravity.

Page No 50:

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,

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

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