Ncert Solution Class 12th Physics Chapter – 1 Electric Charges and Fields
| Textbook | NCERT |
| class | Class – 12th |
| Subject | Physics |
| Chapter | Chapter – 1 |
| Chapter Name | Electric Charges and Fields |
| Category | Class 12th Physics Question & Answer |
| Medium | English |
| Source | last doubt |
Ncert Solution Class 12th Physics Chapter – 1 Electric Charges and Fields
?Chapter – 1?
✍Electric Charges and Fields✍
?Question & Answer?
Q 1. What is the force between two small charged spheres having charges of 2 × 10–7C and 3 × 10–7 C placed 30 cm apart in the air?
Answer:
Q 2. The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in the 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,
Where, ∈0 = Permittivity of free space
The distance between the two spheres is 0.12 m.
(b) Since the spheres have opposite charges, the force on the second sphere due to the first sphere will also be equal to 0.2N.
Q 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
This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.
Q 4. (i) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(ii) 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.
Q 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: When two bodies are rubbed against each other, a charge is developed on both bodies. These charges are equal but opposite in nature. And this phenomenon of inducing a charge is known as charging by friction. The net charge on both of the bodies is 0 and the reason behind it is that an equal amount of charge repels it. When we rub a glass rod with a silk cloth, charge with opposite magnitude is generated over there. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.
Q 6. Four point charges q_{A} = 2 \mu C, \; q_{B} = -5 \mu C, \; q_{C} = 2 \mu C, \; and q_{D} = -5 \mu C 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.
Where, (Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD = 
cm
AO = OC = DO = OB = 
cm
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.
Q 7. (i) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
(ii) Explain why two field lines never cross each other at any point.
Answer: (i) When a charge is placed in an electrostatic field then it experiences a continuous force. Therefore, an electrostatic field line is a continuous curve. And a charge moves continuously and does not jump from on point to the other. So, the field line cannot have a sudden break.
(ii) if two field lines will cross each other at any point then at that point the field intensity will start shooing two directions at the same point which is impossible. Therefore, two field lines can never cross each other.
Q 8. Two point charges qA = 3 µC and qB= –3 µC are located 20 cm apart in a vacuum.
(i) What is the electric field at the midpoint O of the line AB joining the two charges?
(ii) 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.
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 = 
along OB
Where,
∈o = Permittivity of free space
1/4π ∈o = 9 × 10 o N m2 C-2
Magnitude of electric field at point O caused by −3μC charge,
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.
(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.
Q 9. A system has two charges q _{A} = 2.5 \times 10 ^{-7} \;C \; and \; q _{B} = -2.5 \times 10 ^{-7}\;C located at points A :(0, 0, -15 cm) and B (0, 0, + 15 cm), respectively. What is the total charge and electric dipole moment of the system?
Answer: According to the figure given below both the charges can be located in a coordinate frame of reference.
At A, amount of charge, qA = 2.5 × 10−7C (Given)
At B, amount of charge, qB = −2.5 × 10−7 C (Given)
Total charge of the system,
q = qA + qB = 2.5 × 10-7 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.
Q 10. An electric dipole with dipole moment 4 \times 10^{-9}\; C\, m is aligned at 30° with the direction of a uniform electric field of magnitude 5 \times 10^{4}\; N C^{-1}. Calculate the magnitude of the torque acting on the dipole.
Answer: Electric dipole moment, p = 4 × 10−9 C m (Given)Angle made by p with a uniform electric field, θ = 30° (Given)
Electric field, E = 5 × 104 N C−1 (Given)
Torque acting on the dipole is given by the relation,
τ = pE sin θ
= 4 × 10-9 × 5 × 104 × sin 30
= 20 × 10-5 × 1/2
= 10-4 Nm
Therefore, the magnitude of the torque acting on the dipole is 10−4 N m.
Q 11. A polythene piece rubbed with wool is found to have a negative charge of 3 \times 10^{-7}\; C.
(i) Estimate the number of electrons transferred (from which to which?)
(ii) 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
n can be calculated using the relation, q = ne
n = q/e
= -3 × 10-7 / -1.6 × 10-19
= 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 occur. This is because an electron has mass,
as we know
me = 9.1 × 10−3 kg
Total mass transferred from wool to polythene,
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.
Q 12. (i) Two insulated charged copper spheres A and B have their centres
separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 \times 10^{-7} \; C each? The radii of A and B are negligible compared to the distance of separation.
(ii) 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,
f = qA qB / 4π ∈o r2
Where, ∈0 = Free space permittivity
1 / 4π∈o
=9×109N m2 C2
∴ f = 9 × 109 × (6.5 ×10-7)2 / (0.5)2
=1.52X10-2N Therefore, the force between the two spheres is 1.52 × 10−2 N.
(b) As given in the question
After doubling the charge of sphere,
charge on sphere A would be,
qA = 2 × 6.5 × 10−7 C
and charge on sphere B would be,
qB = 1.3 × 10−6 C
The distance between the spheres is halved as given:
∴ r = 0.5 / 2 = 0.25 m
Now,force of repulsion between the two spheres,
f = qA qB / 4π ∈o r2
= 9 × 109 × 1.3 ×10-6 × 1.3 × 10 -6 (0.25)2
= 16 × 1.52 × 10−2
= 0.243 N
Therefore, the force of repulsion between the two spheres is 0.243 N.
Q 13. Suppose the spheres A and B in Exercise 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 mInitially, the charge on each sphere, q = 6.5 × 10 −7 C
When sphere A is touched with an uncharged sphere C, q/2 amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, is q/2.
When sphere C with charge q/2 is brought in contact with sphere B with charge q, total charges on the system will divide into two equal halves given as,
1/2 (q + q/2) = 3q/4
Hence, charge on each of the spheres, C and B, is 3q/4.
Force of repulsion between sphere A having charge q/2 and sphere B having 3q/4 charge is
Therefore, the force of attraction between the two spheres is 5.703 × 10 −3 N.
Q 14. The figure below 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.Q 15. Consider a uniform electric field E = 3 × 10 3 î N/C.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the y z – 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, E = 3 × 10 3 î N / C
Magnitude of electric field intensity, | E | = 3 × 10 3 N / C
Side of the square, s = 10 cm = 0.1 m
Area of the square, A = s 2 = 0.01 m 2
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 = | E |A cos θ
φ = 3 × 10 3 × 0.01 × cos 0 °
φ = 30 N m 2 /C
(b) Plane makes an angle of 60 ° with the x – axis.
Hence, θ = 60°
Flux, φ = | E |A cos θ
Flux, φ = 3 × 10 3 × 0.01 × cos 60°
Flux, φ = 30 x 0.5
Flux, φ = 15 N m 2 /C
Q 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.
Q 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 × 10 3 N m 2 /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,
ø = q / ∈O
∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−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.
Q 18. A point charge + 10 μC is at 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,
∅ = ∅Total/6
= 1/6 q/∈o
Where, ∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2 q = 10 μC = 10 × 10−6 C
∴ = 1/6 × 10 ×10-6 / 8.854 × 10-12
= 1.88 × 105 N m2 C−1
Therefore, electric flux through the square is 1.88 × 105 N m2 C−1.
Q 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,∅Net = q/∈o
Where, ∈0 = Permittivity of free space
= 8.854 × 10−12 N−1C2 m−2
q = Net charge contained inside the cube = 2.0 μC = 2 × 10−6 C
∴ ∅Net = 2 × 10-6 /8.854 × 10−12
= 2.26 × 105 N m2 C−1
The net electric flux through the surface is 2.26 ×105 N m2C−1.
Q 20. A point charge causes an electric flux of – 1.0 × 10 3 N m 2 /C to pass through a spherical Gaussian surface of 10.0 cm radius centred 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/CRadius 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,
∅ = q/∈o
Where, q = Net charge enclosed by the spherical surface
∈0 = Permittivity of free space = 8.854 × 10−12 N−1C2 m−2
∴ ∅ = q/∈o
= −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.
Q 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 × 10 3 N/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,E = q/4π ∈o
Where, q = Net charge = 1.5 × 103 N/C
d = Distance from the centre = 20 cm = 0.2 m
∈0 = Permittivity of free space
And, 1/4π∈o
= 9 × 109 N m2 C−2
∴ q = E(4π ∈o d2
= 1.5 × 103 × (0.2)2 9 × 109
= 6.67 × 109 C
= 6.67 nC
Therefore, the net charge on the sphere is 6.67 nC.
Q 22. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC /m 2
(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 mRadius of the sphere, r = 1.2 m
Surface charge density, σ = 80.0 μC/m2 = 80 × 10−6 C/m2
Total charge on the surface of the sphere,
Q = Charge density × Surface area
= σ × 4πr2
= 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 (∅Total) leaving out the surface of a sphere containing net charge Q is given by the relation,
∅Total = Q/∈o
Where, ∈0 = Permittivity of free space
∈0 = 8.854 × 10−12 N−1C2 m−2
Q = 1.447 × 10−3 C
∅Total = 1.44 × 10-3/8.854 × 10-12
= 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.
Q 23. An infinite line charge produces a field of 9 × 10 4 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,E = λ/2π ∈o d
λ = 2π ∈o dE
Where,
d = 2 cm = 0.02 m
E = 9 × 104 N/C
∈0 = Permittivity of free space
1/4π ∈0 = 9 × 109 N m2 C−2
λ = 0.02 × 104 / 2 × 9 × 109
= 10 μC/m
Therefore, the linear charge density is 10 μC/m.
Q 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 /m 2
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,
E = σ/∈o
Where,
∈0 = Permittivity of free space = 8.854 × 10−12 N−1C2 m−2
∴ E = 17.0 × 10-22 / 8.854 × 10-12
= 1.92 × 10−10 N/C
Therefore, electric field between the plates is 1.92 × 10−10 N/C.
Q 25. An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 NC–1 (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm–3. Estimate the radius of the drop.
(g = 9.81 m s–2; e = 1.60 × 10–19 C).
Answer: Excess electrons on an oil drop, n = 12Electric 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 = 4/3πr3 × ρ × g
Where, q = Net charge on the oil drop = ne
m = Mass of the oil drop
= Volume of the oil drop × Density of oil
= 4/3πr3 × ρ
= 9.82 × 10−4 mm
Therefore, the radius of the oil drop is 9.82 × 10−4 mm.
Q 26. Which among the curves shown in Fig. cannot possibly represent electrostatic field lines?
Answer: (a) The field lines are not normal to the surface of the conductor. Therefore, it does not represent electrostatic field lines.
(b) In this figure, the field lines are emerging from the negative charge and terminating at the positive charge. This is not possible. Therefore, it does not represent electrostatic field lines.
(c) The electric field is emerging from the positive charge and it repels each other. Therefore, the field lines shown represent electrostatic field lines.
(d) The field lines shown intersect each other. Therefore, the field lines shown does not represent electrostatic field lines.
(e) The closed loops are formed in the area between the field lines. Therefore, it does not represent electrostatic field lines.
Q 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: Total dipole moment of the system, p=q×dl=−10−7 cm
The rate of increase of the magnitude of the electric field along the positive z-direction = 105 NC–1 per metre.
The force experienced by the system is given as
F=qE
F=q(dE/dl) x dl
= p x (dE/dl)
=- 10−7 ×10−5
=−10−2N
The force experienced by the system is – 10−2 N. This is in the negative z-direction and opposite to the direction of the electric field. Therefore, the angle between the dipole moment and the electric field is 180o.
Torque (τ) =pEsin1800=0
Therefore, the torque experienced by the system is zero.
Q 28. (a) A conductor A with a cavity as shown in Fig. (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 q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Fig. (b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
Answer: (a) A Gaussian surface is considered within the conductor that encloses the cavity. Within the conductor, the electric field intensity (E) will be zero.
The charge inside the conductor is q and the permittivity of free space is ∈
According to Gauss’s law,
Flux, ϕ=E.ds = q/∈0
Electric field intensity, E=0
q/∈0 = 0
The permittivity of the free space, ∈0 ≠ 0
Therefore, the charge inside the conductor, q=0
Therefore, the entire charge appears on the outer surface of the conductor.
(b) A conductor B is kept in the cavity and it is insulated from A. Hence a charge – q will be induced in the inner surface of conductor A, this in turn will induce a charge +q on the outer surface of A. Therefore, the total charge on the outer surface of A will be equal to Q+q.
(c) The sensitive instrument can be shielded from the electrostatic field by keeping it fully enclosed inside a metallic surface.
Q 29. A hollow charged conductor has a tiny hole cut into its surface. Show that the σ/2ε0 n̂ , where n̂ is the unit vector in the outward normal direction and σ 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, σ is the charge density and ε 0 is the permittivity of free space.
Charge q = σ × ds
According to Gauss’s law, flux, ∅ = E. ds = q/ε0
⇒ E. ds = σ × ds / ε0
∴ E= σ/ 2ε0 n̂
Therefore, the electric field just outside the conductor is σ/ 2ε0 n̂. This field is a superposition of field due to the cavity E` and the field due to the rest of the charged conductor E`. These fields are equal and opposite inside the conductor and equal in magnitude and direction outside the conductor.
∴ E`+ E` = E
⇒ E` = E/2 = σ/2ε0 n̂
Hence, the field due to the rest of the conductor is σ/2ε0 n̂.
Q 30. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E 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 following figure) of uniform linear charge density λ.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.
∴ q = λdx
Electric field due to the piece,
However,
The electric field is resolved into two rectangular components. dEcosθ is the perpendicular component and dEsinθ is the parallel component. When the whole wire is considered, the component dEsinθ is cancelled. Only the perpendicular component dEcosθ affects point A.
Hence, effective electric field at point A due to the element dx is dE1 .
∴ d E1 = 1/4πε0 x λdx. cos θ/(l2 + x2) …(1)
In ∆AZO, tanθ = x/l ⇒ x = l.tanθ …(2)
On differentiating equation (2), we obtain
dx/dθ = l x sec 2 θ ⇒ dx = l x sec 2 θdθ …(3)
From equation (2), we have
x2 + l2 = l2tan2θ + l2 = l2 (tan2 θ + 1) = l2 sec2 θ …(4)
Putting equations (3) and (4) in equation (1), we obtain
The wire is so long that θ tends from − π/2 to π/2.
By integrating equation (5), we obtain the value of field E1 as,
Therefore, the electric field due to long wire is λ/2πε0 l .
Q 31. It is now established 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 (+2/3) e.
Charge due to n up quarks = (2/3 e) n
Number of down quarks in a proton = 3 − n
Each down quark has a charge of – 1/3e
Charge due to (3 − n) down quarks = (-1/3 e) (3-n)
Total charge on a proton = + e
Therefore, e = (2/3 e) n + (- 1/3 e) (3-n)
⇒ e = 2ne/3 – e + ne/3
⇒ 2e = ne ⇒ n = 2
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.
Charge on a neutron due to n up quarks = (+ 2/3 e) n
Number of down quarks is 3 − n, and each having a charge of – 1/3 e.
Charge on a neutron due to (3 – n) down quarks = (- 1/3 e) (3-n)
Total charge on a neutron = 0
Therefore, 0 = (2/3 e) n + (-1/3 e) (3-n)
⇒ 0 = 2ne/3 – e + ne/3
⇒ e = ne ⇒ n = 1
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’.
Q 32. (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 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 us consider that the equilibrium is stable. Then if the test charge is displaced in any direction it will experience a restoring force towards the null-point. This means that all the field lines near the null point will be directed towards the null point. That is, there is a net inward flux of electric field through a closed surface around the null-point. But according to Gauss’s law, the flux of electric field through a surface, not enclosing any charge, must be zero. Hence, equilibrium cannot be stable.
(b) The null-point is the mid-point of the line joining the two charges. If the test charge is displaced along the line from the null-point there is a restoring force. If the test charge is displaced normal to the line the net force takes it away from the null-point. Hence for the stability of equilibrium needs restoring force in all directions.
Q 33. A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed vx (like the figure in question 1.14). The length of the plate is L and a uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL2/(2m vx2). Compare this motion with the motion of a projectile in a gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.
Answer:
Q 34. Suppose that the particle in Exercise 1.33 is an electron projected with velocity vx = 2.0 × 106 m s–1. If E between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate? (|e|=1.6 × 10–19 C, me = 9.1 × 10–31 kg.)
Answer:
