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
CBSE Guide NCERT Solutions - Download Class12 CBSE Physics NCERT Solutions of Electric Charges and Fields




(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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