NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields: Electrostatics is the study of electric charges at rest. The chapter electric charges and fields deal with the charging of a body, properties of charge, Columbus law, electric field, electric flux, Gauss law and application of Gauss law. Solutions of NCERT class 12 physics chapter 1 electric charges and fields explain problems related to these topics. Try solving all the questions of NCERT and if any doubt arises then check the CBSE NCERT solutions for class 12 physics chapter 1 electric charges and fields. The derivations in NCERT chapter 1 electric charges and fields are very important in CBSE board exam point of view. Also, the steps in derivations are important as it will help in NCERT solutions for class 12 physics chapter 1 electric charges and fields. As the second chapter of NCERT class 12 is the continuation of the first chapter, so it is important to remember all points studied in the chapter. The NCERT solutions will help you in better understanding of the concepts you have studied. While studying NCERT solutions for class 12 physics chapter 1, electric charges and fields you can compare the formulas of this chapter with the chapter gravitation of class 11 NCERT.
Electrostatics and gravitation some comparisons
The force between two charges separated by a distance r
The force between two masses separated by a distance r
NCERT solutions for class 12 physics chapter 1 electric charges and fields exercise:
Q 1.3 Check that the ratio is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
So the dimension of
The gravitational force between two bodies of mass M and m is
so dimension of
Therefore from (1) and (2)
K = , where is the permittivity of space.
e = Electric charge ([e] = [C])
G = Gravitational constant. ([G]= )
and are mass of electron and proton ([ ] = [ ] = [Kg])
Substituting these units, we get
Hence, this ratio is dimensionless.
Putting the value of the constants
The given ratio is the ratio of electric force to the gravitational force between an electron and a proton considering the distance between them is constant!
Q 1.4 (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.
The given statement "electric charge of a body is quantised" implies that charge on a body can take only integral values. In other words, only the integral number of electrons can be transferred from one body to another and not in fractions.
Therefore, a charged body can only have an integral multiple of the electric charge of an electron.
Q 1.7 (b) Explain why two field lines never cross each other at any point?
A tangent drawn at any point on a field line gives the direction of force experienced by a unit positive charge due to the electric field on that point. If two lines intersect at a point, then the tangent drawn there will give two directions of force, which is not possible. Hence two field lines cannot cross each other at any point.
Q 1.8 (a) Two point charges and 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?
Given, AB = 20 cm
Since, O is the midpoint of the line AB.
AO = OB = 10 cm = 0.1m
The electric field at a point caused by charge q, is given as,
Where, q is the charge, r is the distance between the charges and the point O
k = 9x10 9 N m 2 C -2
Due to charge at A, electric field at O will be and in the direction AO.
Similarly the electric field at O due to charge at B, also in the direction AO
Since, both the forces are acting in the same direction, we can add their magnitudes to get the net electric field at O:
E' = + = 2E (Since their magnitudes are same)
along the direction AO.
Q 1.9 A system has two charges and 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?
The total charge of the system =
The system is electrically neutral. (All dipole systems have net charge zero!)
Now, distance between the two charges, d = 15 + 15 = 30 cm = 0.3 m
We know, The electric dipole moment of the system, p = x d = x d (i.e, the magnitude of charge x distance between the two charges)
The direction of a dipole is towards the positive charge . Hence, in the positive z-direction.
Using Gauss’s law, we know that the flux of electric field through any closed surface S is times the total charge enclosed by S.
where, q = net charge enclosed and = permittivity of free space (constant)
Therefore, flux does not depend on the radius of the sphere but only on the net charge enclosed. Hence, the flux remains the same although the radius is doubled.
We know, for determining the electric field at r>R for a conducting sphere, the sphere can be considered as a point charge located at its centre.
Also, electric field intensity at a point P, located at a distance r, due to net charge q is given by,
Given, r = 20 cm = 0.2 m (From the centre, not from the surface!)
Therefore, charge on the conducting sphere is (since flux is inwards)
Using Gauss's law, we know that :
Q 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 σ is the surface charge density near the hole.
Let the surface area of the sphere be S.
And assume that the hole is filled. For a point B, just above the hole, considering a gaussian surface passing through B, we have
Now, since the electric field is always perpendicular to the surface of the conductor.
Using Superposition principle, ,
where is due to the hole and is due to the rest of the conductor. (both pointing outwards, i.e away from the centre)
Again, for a point A, just below the hole, Electric field will be zero because of electrostatic shielding.
Using the superposition principle, this will be due to pointing inwards(towards the centre) and due to (Pointing away from the centre)
Using this relation, we get:
Since this is pointing outwards,
is the electric field in the hole.
(Trick: 1. Assume the hole to be filled.
2. Consider 2 points just above and below the hole.
3. Electric fields at these points will be due to the hole and rest of the conductor. Use superposition principle.)
Q 1.30 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.]
Let AB be a long thin wire of uniform linear charge density λ.
Let us consider the electric field intensity due to AB at point P at a distance h from it as shown in the figure.
The charge on a small length dx on the line AB is q which is given as q = λdx.
So, according to Coulomb’s law, the electric field at P due to this length dx is
This electric field at P can be resolved into two components as dEcosθ and dEsinθ. When the entire length AB is considered, then the dEsinθ components add up to zero due to symmetry. Hence, there is only dEcosθ component.
So, the net electric field at P due to dx is
dE' = dE cosθ
⇒ x = h tan θ
Differentiating both sides w.r.t. θ,
⇒ dx = h sec 2 θ dθ …………………….(2)
Also, h 2 + x 2 = h 2 + h 2 tan 2 θ
⇒ h 2 + x 2 = h 2 (1+ tan 2 θ)
⇒ h 2 + x 2 = h 2 sec 2 θ ………….(3)
(Using the trigonometric identity, 1+ tan 2 θ = sec 2 θ)
Using equations (2) and (3) in equation (1),
The wire extends from to since it is very long.
Integrating both sides,
This is the net electric field due to a long wire with linear charge density λ at a distance h from it.
In the question linear charge density =E
Q 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 + e, and the ‘down’ quark (denoted by d) of charge 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.
Given, a proton and a neutron consist of three quarks each.
And, ‘up’ quark is of charge + e, and the ‘down’ quark of charge e
Let the number of 'up' quarks be n. Therefore, the number of 'down' quarks is (3-n).
The net charge =
Now, a proton has a charge +1e
Proton will have 2 u and 1 d, i.e, uud
Similarly, the neutron has a charge 0
Neutron will have 1 u and 2 d, i.e, udd
Q 1.33 A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed vx .The length of plate is L and an 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 .
Let s be the vertical deflection, t be the time taken by the particle to travel between the plates
Here , u =0 , since initially there was no vertical component of velocity.
The particle in the elecric field will experience a constant force (Since, Electric field is constant.)
F = ma = -qE (Using Newton's Second Law, F = ma)
a = -qE/m (-ve sign implies here in downward direction)
Again, t = Distance covered/ Speed =
(In x-direction, since there is no force, hence component of velocity in x-direction remains constant = .
And, the distance covered in x-direction = length of the plate = L)
Putting these values in our deflection equation,
(S is -ve, which implies it deflect in downwards direction.)
The vertical deflection of the particle at the far edge of the plate is .
This motion is similar to the motion of a projectile in a gravitational field, which is also a constant force. The force acting on the particle in the gravitational field is mg whiles in this case, it is qE. The trajectory will be the same in both cases.
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Importance of NCERT solutions for class 12 physics chapter 1 electric charges and fields in board exams:
The physics paper for CBSE board exam is for 70 marks. In 2019 CBSE board exam 9 marks question were asked from the session electrostatics that includes the first two chapters of NCERT class 12 physics. Practising problems from the NCERT solutions for class 12 physics chapter 1 electric charges and fields help students to score well in the final exam.