Cbse Ncert Solutions of Class 12 Physics: Electric Charges and Fields (Q 17-24)


Chapter 1, Electric Charges and Fields

NCERT Solutions of 12th CBSE Physics Textbook Exercises


Question 1.17: Careful measurement of the electric field at the surface of the black box indicates that the net outward flux through the surface of the box is 8.0 x 103 Nm2/C
(i) What is the net charge inside the box?
(ii) 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:
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(ii) We cannot conclude that the net electric charge inside the box is zero if the outward flux through the surface of black box is zero because there might be equal amounts of positive and negative charges cancelling each other and thus making the resultant charge equal to zero. Thus, we can only conclude that the net charge inside the box is zero.
Question 1.18: A point charge + 10µC is a distance 5 cm directly above the centre of a square of side 10 cm as shown in figure. 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 given square ABCD can be imagined as one of the side faces of a cube of side 0.10 m. The given charge can be imagined to be at the centre of this cube at a distance of 5 cm.
Here, q = + 10µC = 10-5 C.
Then, according to the Gauss’s Theorem, the total electric flux through all the 6 faces of the cube is given by -
https://www.cbsencertsolution.com - Cbse Ncert Solutions of Class 12 Physics: Electric Charges and Fields textbook question 1.18

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: Here, charge at the centre of the Gaussian surface,
q = 2µC = 2 x 10-6 C.
ɛo = 8.854 x 10-12 C2 N-1 m-2
Ф = electric flux through it = ?
According to Gauss’s Theorem, the electric flux through the six faces of the cubes i.e., through the Gaussian surface is given by -
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Question 1.20: A point charge causes an electric flux of -1.0 x 103 Nm2/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: Here, Ф = electric flux through the spherical Gaussian surface = -1.0 x 103 Nm2 C-1.
r = radius of Gaussian spherical surfaces = 10 cm.
Let, q = charge enclosed at its centre.
(a) According to Gauss’s law, the electric flux through a Gaussian surface depends upon the charge enclosed inside the surface and not upon its size. Thus, the electric flux will remain unchanged i.e., -1.0 x 103 Nm2 C-1 through the spherical Gaussian surface of double radius i.e. of 20 cm as it also encloses the same amount of charge.
(b) q = point charge = ?, ɛo= 8.854 x 10-12 Nm-2C2.





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 x 103 N/C and points radially inward, what is the net charge on the sphere ?
Answer: Here, R = radius of the conducting sphere = 0.10 m
r = distance of the point from centre of sphere = 20 cm = 0.2 m
Clearly r > R
E = electric field at a point at a distance of 20 cm from the sphere = 1.5 x 103 NC-1 acting inward.
q = net charge on the sphere = ?
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: Here, σ = surface charge density = 80.0 µC/m2= 80 x 10-6 C/m2
    R = radius of the charged sphere = 2.4/2 = 1.2 m.
(a)  q = charge on the sphere = ?
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Question 1.23: An infinite line charge produces a field of 9 x 104 N/C at a distances of 2 cm. Calculate the linear charge density.
Answer: Here, E = electric field produced by infinite line charge = 9 x 104 NC-1.
r = distance of the point where E is produce = 2 cm = 0.02 m.
λ = linear charge density = ?
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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 x 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, (c) between the plates?
Answer:
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The arrangement of the plates is shown in the figure where, σ = surface density of plates = 17.0 x 10-22 Cm-2, ɛo = 8.854 x 10-12 C2 N-1m-2.
(a) E in the outer region of the first plate = E to the left of the first plate = ? The region I is to the left of the first plate thus, electric field E1 in this region due to the two plates is given by the following equation:
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(b) E in the outer region of second plate = E to the right of second plate, i.e., in region III =?. The E for the region III is given by -
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(c) E between the plates = E in II region = EII =? EII is given by -
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 Class XII CBSE Physics - Chapter 1, Electric Charges and Fields - NCERT Solutions


# CBSE Physics Guide - Electric Charges and Fields - Study materials, Additional Objective Questions, Short Questions & Long Questions, Numerical Problems with Solutions important for CBSE 12th Board Exam



Download 12th CBSE Physics NCERT Answers of Chapter 1 Electric Charges and Fields (Q 9-16)

Chapter 1, Electric Charges and Fields

NCERT Answers of 12th CBSE Physics Textbook Exercise Questions

Question 1.9: A system has two charges qA = 2.5 x 10-7 C and qB = -2.5 x 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: The charges qA and qB are located at points A (0, 0, -15 cm) and B (0, 0, +15 cm) on z –axis as shown in the figure here. They form an electric dipole, where,
q = total charge =?
p = electric dipole moment of the system =?
p = q.2a
Now,
q = qA + qB = 2.5 x 10-7 + (-2.5 x 10-7) = 0
Also, 2a = AB = dipole length = OA + OB
= 15 + 15 = 30 cm = 0.30 m.
Hence, p = either charge x dipole length
               = qA x AB = 2.5 x 10-7 x 0.30
               = 7.5 x 10-8 cm.
The electric dipole moment is directed from B to A i.e., along negative z-axis.
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Question 1.10: An electric dipole with dipole moment 4 x 10-9 C m is aligned at 300 with the direction of a uniform electric field of magnitude 5 x 104 NC-1. Calculate the magnitude of the torque acting on the dipole.
Answer: Here, p= 4 x 10-9 Cm, E = 5 x 104 NC-1, θ = 300, t =?
Using the formula, t = pE sinθ, we get
t  = 4 x 10-9 x 5 x 104 x sin 300
   = 20 x 10-5 x ½
   = 10-4 Nm.
Question 1.11: A polythene piece rubbed with wool is found to have a negative charge of 3 x 10-7C.
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene ?
Answer: (a) Here, q = total charge transferred = -3 x 10-7 C. Charge on an electron, e = -1.6 x 10-19 C.
n = no. of electrons transferred =?
As the polythene piece rubbed with wool is found to attain –ve charge, so the electrons are transferred from wool to polythene piece.
From quantisation of charge, we know that q = ne
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(b) Yes, there is a transfer of mass from wool to polythene as electrons are material particles and are transferred from wool to polythene piece.
m = mass of each electron = 9.1 x 10-31 kg, n = 1.875x 1012
M = total mass transferred to polythene =?
    = m x n = 9.1 x 10-31 x 1.875 x 1012 = 1.71 x 10-18 kg or, 2 x 10-18 kg.
Question 1.12: (a) 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 x 10-7 C? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion of each sphere is charged double the above amount, and the distance between them is halved?
Answer: (a) Charge on 1st sphere, A = qA = 6.5 x 10-7 C
Charge on 2nd sphere, B = qB = 6.5 x 10-7 C
Distance between sphere A and B = 50 cm = 0.5 m = d
We know that,
NCERT Solutions of Class 12 CBSE Physics - Electric Charges and Fields








(b) If each sphere is charged double the amount, then,
qA = qB = 2 x 6.5 x 10-7 C = 13 x 10-7 C
and, r = 1/2 x 50 cm = 0.25 m.
We know that,
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Question 1.13: Suppose the spheres A and B of previous exercise have identical sizes. A third sphere of the same size but unchanged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the force of repulsion between A and B ?
Answer: Initial charges on spheres A and B are qA = qB = 6.5 x 10-7 C
r = Distance between the spheres A and B = 0.5 m.
All the spheres will have equal charges on being brought in contact because they are of the same size.
When 3rd sphere C having charge qc = zero is brought in contact with 1st sphere A, then;
Charge on A = Charge on C = q’1 say

























Question 1.14: Figure given 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 ?
Download NCERT Answers of Electric Charges and Fields, Class 12 CBSE Physics

Answer












Question 1.15: Consider a uniform electric field E = 3 x 103 i 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 yz plane ? (b) What is the flux through the same square if the normal to its plane makes at 60o angle with the x-axis ?
Answer




























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: Net flux over the cube is zero because the number of lines entering the cube of side 20 cm is same as the number of lines leaving the cube.

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Download NCERT Solutions of Class 12 CBSE Physics Chapter 1, Electric Charges and Fields (Q 1-8)

Electric Charges and Fields

NCERT Solutions of Class XII, CBSE Physics Chapter 1 Textbook Exercise Questions

Question 1.1: What is the force between two small charged spheres having charges of 2 x 10-7 C and 3x10-7 C placed 30 cm apart in air ?
Answer: Here, q1 = 2x10-7 C, q2 = 3x10-7 C, r = 0.30m, F =?
Using the relation,
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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: Given that,
q1 = 0.4 µC = 0.4x10-6 C, q2 = -0.8 µC = -0.8x10-6 C,
F= electrostatic force between q1 and q2 = 0.2 N.
(a) r=?
Using the relation,
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(b) Force on q2 due to q1 =?
We know that electrostatic forces always, appear in priors and follow Newton’s 3rd law of motion.
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Question 1.3: Check that the ratio ke2 / Gmemp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does this ratio signify ?
Answer:
Dimensions of e2 are = [C2]
Dimensions of k are = [Nm2 c2] = [ML3 T-2 C-2]
Dimensions of G are = [M-1 L3 T-2]
Dimensions of me = [M]
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Question 1.4: (a) Explain the meaning of the statement ‘electric charge of a body is quantized’.
(b) Why can one ignore quantization of electric charge when dealing with macroscopic i.e., large scale charges?
Answer:
(a) The meaning of the statement ‘electric charge of a body is quantized’ is that the charge on it is always some integral multiple of elementary charge of an electron or a proton (= e in magnitude) i.e., discrete packets called quanta or packets of charge. Mathematically, the charge on a body can be expressed as -
q = ± ne where,
n = an integer, e = magnitude of the charge of an electron or proton = 1.6x10-19 C.
A fraction of the fundamental charge e has never been observed in free state.

(b) In practice, the charge on a charged body at macroscopic level is very large while, the charge on an electron is very small. When electrons are added to or removed from a body, the change taking place in the total charge on the body is so small that the charges seems to be varying in a continuous manner. Thus, quantization of electric charge can be ignored at macroscopic level i.e., when dealing with a large scale charged body.

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 other law of conservation of charge.
Answer: Initially both the glass rod and silk cloth are electrically neutral before rubbing. In other words, net charge on the glass rod and silk cloth is zero. When the glass rod is rubbed with silk cloth, thus, glass rod becomes positively charged and silk cloth negatively charged. The positive charge on the glass rod is exactly equal to the negative charge on the silk cloth, so net charge on the system is again zero. Thus, the appearance of charge on the glass rod and silk cloth is in accordance with the law of conservation of charge as the total charge of the isolated system is constant. Similarly, when ebonite rod is rubbed with fur, they acquire –ve and +ve charges respectively and net charge is zero again.
Thus, we conclude that charge is neither created nor destroyed but it is merely transferred from one body to another which is consistent with the law of conservation of charge.

Question 1.6: Four points 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 ?
Answer: Consider the square ABCD of each side 10 cm and centre O. The charge of 1 µC is placed at O. Now clearly,
OA = OB = OC = OD
AB = BC = 10 cm = 0.1 m
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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) The electrostatic line of force is the path tangent at every point of which gives the direction of electric field at that point. The direction of electric field generally changes from point to point. So, the lines of force are generally curved lines. Further, they are continuous curves and cannot have sudden breaks if it is so, then it will indicate the absence of electric field at the break points.

b) The electric lines of force never cross each other because if they do so, then, at the point of their intersection, we can draw two tangents which give two directions of electric field at that point which is not possible.
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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 mid-point O of the line AB joining the two charges ?
(b) If a negative test charge of magnitude 1.5 x 10-29 C is placed at this point, what is the force experienced by the test charge ?
Answer:
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Here, charge at point A, qA = +3 µC = 3x10-6 C. 
Charge at point B, qB = -3 µC = -3x10-6 C.
r = AB = 20 cm = 0.2 m.
Let O be the mid-point of the line AB, then,
OA = OB = r/2 = 0.2/2 = 0.1 m.
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(b) Force on a negative charge of magnitude 1.5 x 10-9 C is given by the formula.
https://www.cbsencertsolution.com/2018/08/download-ncert-solutions-of-class-12-cbse-physics-chapter-1-electric-charges-and-fields.html













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