NCERT Solutions for Class 12 Physics, Chapter 1: Electric Charges and Fields are available here. Due to a large number of questions, we have divided the solutions of this chapter into three parts. This is Part 1, it contains solutions from question number 1.1 to question number 1.11. Solutions of other problems are available in further parts.
NCERT Solutions for Class 12 Physics, Chapter 1: Electric Charges and Fields (Part 1) are given below:
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:
Q_{1}_{ }= 2 × 10^{-7} C
Q_{2} = 3 × 10^{-7} C
r = 30 × 10^{-2} m
We know,
F = (Q_{1} Q_{2})/(4 π ε_{o} r^{2} )
Substituting the given values, we get
F = 6 × 10^{-3} N.
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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 = (Q_{1} Q_{2})/(4πε_{o} r^{2} )
Substituting the values, we get
r = 0.12 m = 12 cm.
(b) Force on second sphere due to first
F_{21} = (Q_{1} Q_{2})/(4πε_{o} r^{2} )
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 {ke^{2}} /{G m_{e} m_{p}} 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 × 10^{9} Nm^{2} C^{‒2}
e = 1.6 × 10^{‒19} C
G = 6.6 × 10^{‒11} Nm^{2} kg^{‒2}
m_{e} = 9.11 × 10^{‒31} kg
m_{p} = 1.6 × 10^{‒27} kg
Putting these values in{ke^{2}} /{G m_{e} m_{p}} its value is 2.29 × 10^{39}.
Ratio = {ke^{2}} /{G m_{e} m_{p}} = {[Nm^{2} C^{‒2}] [C]}/{[Nm^{2} kg^{‒2}] [kg] [kg]} = 1
Hence, the given ration is dimensionless.
The ratio signifies that electrical forces are immensely stronger than the gravitational forces.
CBSE Class 12 Physics Syllabus 2017-18
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., ne, n = …-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 q_{A} = 2 μC, q_{B} = –5 μC, q_{C} = 2 μC, and q_{D} = –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_{A }= 3 μC and q_{B} = –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:
q_{A}_{ }= 3 μC = 3 × 10^{‒6} C
q_{B} = ‒3 μC = ‒ 3 × 10^{‒6} C
r = 10 cm (point O is in middle)
We know, electric field is given by formula
E = Q/(4πϵ_{o} r^{2} )
Electric field due to charge A (by using above formula)
|E_{A}| = 2.7 × 10^{6} N / C
Electric field due to charge B
|E_{B}| = 2.7 × 10^{6} N/C opposite to the direction of E_{A}_{.}
Since both the fields have the same direction, the electric field at O will be summation of above field.
So, |E_{Net}| = 5.4 × 10^{6} N/C towards B.
(b) Given:
q_{o} = 1.5 × 10^{–9} C
|E_{Net}| = 5.4 × 10^{6} N/C towards B
By using the formula
F = q E
We have,
F = (q_{o}) (|E_{Net}|) = (1.5 × 10^{–9} C) × (5.4 × 10^{6} N/C) = 8.1 × 10^{‒3} towards charge q_{B} (because direction of electric field is towards B).
Question1.9: A system has two charges q_{A} = 2.5 × 10^{–7} C and q_{B} = –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 × 10^{4} NC^{–1}. Calculate the magnitude of the torque acting on the dipole.
Solution 1.10:
Dipole moment, |p| = 4 × 10^{-9} Cm, θ = 30^{o}, E = 5 × 10^{4} N/C
We know that,
Torque = p E sinθ
Torque = (4 × 10^{‒9}) (5 ×10^{4}) (sin30^{o})
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, e = 1.6 × 10^{-19} C
Therefore, n = |q|/e = 1.875 × 10^{12} electrons.
1.875 × 10^{12} 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 × 10^{12}) (9.1 × 10^{‒31}) = 1.7 × 10^{‒18 }kg.
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