NCERT Solution Physics Class 12 Chapter 1 Electric Charges and Fields
NCERT Solution Physics Class 12 Chapter 1 Electric Charges and Fields all questions and answers. Physics Class 12 1st Chapter Electric Charges and Fields exercise solution and experts answer. As one of online learning platforms, we (netex.) are excited to offer the NCERT Solution Physics Class 12 Chapter 1. This solution is designed to help students who are looking to brush up on their physics concepts on Chapter 1 Electric Charges and Fields.
Chapter 1 Electric Charges and Fields
1.3)
1.4.) (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?
ANSWER-
The meaning of statement “Electric charge of a body is quantized”. Is that the only integral multiple of smallest unit of charge which is charge on electron (e = 1.6 x 10-19c) can be transferred from one body to the other and they cannot be not transferred in fraction.
In macroscopic view, transfer of charges are very large as compare to smallest unit of charge (e = 1.6 x 10-19c) hence quantisation has no significance in macroscopic scale.
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 law of conservation of charge.
ANSWER-
the law of conservation of charge states that the net charge of isolated system remains conserved. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. We know that individual charge don’t have meaning as charges are produced in pairs hence during Rubbing charges of equal magnitude but of opposite nature are produced on the two bodies and The net charge on the system of two rubbed bodies is zero which is consistent with law of conservation of charge.
1.6)
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)An electrostatic field line is the line of forces which exerts the force on another charge when it comes in field of influence of electrostatic field lines. When charge comes in contact it moves continuously and does not jump from one point to other which proves that field lines are continuous and cannot have sudden break.
(b) we know that tangent to the curve of electric field line represents the direction of electric field intensity. If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.
1.8)
1.10)
1.14) Figure 1.33 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-
We know that same charges repel each other and Opposite charges attract each other. From above figure we can observe 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.
On the other hand particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.
We know that The charge to mass ratio is directly proportional to the amount of deflection for constant velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.
1.15)
1.20.)
In case you are missed :- NCERT Solution for Electrostatic Potential And Capacitance
1.23)
1.26.) Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?
ANSWER-
(a)one of the property of field line is that field lines must be normal to the surface of the conductor. From the figure (a) The field lines are not perpendicular to the conductor hence do not represent electrostatic field lines
(b)one of the property of field line is that field lines are originating from positive charge and goes into negative charge but from figure (b)The field lines are originating from negative charge hence do not represent electrostatic field lines.
(c)One of the property of field line is that field lines are originating from positive charge and goes into negative charge The field lines showed in figure(c) emerges from the positive charges hence represent electrostatic field lines.
(d)One of the properties of field line is thatfield lines not intersect each other. The field lines showed in figure (d) intersects each other hence do not represent electrostatic field lines.
(e)The field lines showed in figure (e) do not represent electrostatic field lines because closed loops are are formed in the figure in the area between the lines which violates property of electric field lines.
1.27)
1.28 (a) A conductor A with a cavity as shown in Fig. 1.36(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. 1.36(b)]. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
ANSWER-
(a)Let us consider a Gaussian surface within a conductor and enclosing the cavity. The electric field intensity (E) inside charged conductor is zero. Using,
Charge inside the conductor is zero hence all the charges will appear on the outer surface of the conductor.
(b)Conductor B with charge q is inserted into the cavity keeping B insulated from A. hence due to this a charge of negative q is attracted and induced on inner surface of conductor A and +q charges get repulted to outer surface of the conductor A. hence total charge on the outer surface will become Q + q.
(c)A sensitive instrument can be shielded from the strong electrostatic field in its environment by placing this instrument inside a metallic surface.
1.29.) A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε0) nˆ, where nˆ 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 hole into its surface. Let E be the electric field outside the conductor then according to gauss law,
If s is surface charge density then,
∴ σ = Charge / Surface area = q/ds
∴ q = σds
This electric field outside the conductor is due to electric field due to cavity and electric field due to rest of the conductor. These fields contributes equally hence the electric field in the hole is ½ E i.e. (σ/2ε0) nˆ, where nˆ is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.
1.30)
1.31.) 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-
It is given that proton has 3 quarks .let consider that there are n number of the up quark with charge + (2/3) e and (3-n) number of down quark with charges (–1/3) e.
∴ nx(2/3) e + (3-n)x(–1/3) e = e
∴ 2/3ne – e + 1/3ne = e
∴ 2/3n – 1 + 1/3n = 1
∴ 2/3n + 1/3n =2
∴ n = 2
Therefore there are two up quarks and one down quark in proton hence the composition of the proton can be written as uud.
We know that for neutron, total charge is zero. It is given that proton has 3 quarks let there are n number of the up quark with charge + (2/3) e and (3-n) number of down quark with charges (–1/3) e.
∴ n x (2/3) e + (3-n) x (–1/3) e = 0
∴ 2/3ne – e + ne = 0
∴ 2/3ne + ne = e
∴ ne = e
∴ n = 1
Therefore for neutron, there are one up quark and 2 down quark hence neutron can be represented by udd.
1.32 (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-
We know that for equilibrium to be stable there should be a net force and hence the net flux towards mean position so that when charge gets displaced from this mean position it will come back to equilibrium position. But as this point is null point ,According to Gauss’s law, the flux ofelectric field through a surface, which is not enclosing any charge, is zero. Hence there is no flux present hence no restoring force to take the displaced charge to equilibrium position hence the test charge isnecessarily in unstable equilibrium.
When Two charges of same magnitude and sign are placed at a certain distance then The mid point of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force towards mean position. If test charge is displaced normal to the line joining them then the net force due to equal charges will produce resultant force in the direction away from the null point hence it gives unstable equilibrium condition.
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 (like particle 1 in Fig. 1.33). 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 qEL2/(2m vx2 ). Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.
ANSWER-
Here particle is moving with the velocity of vx where x is changing hence the velocity is variable which gives rise to the acceleration. The force responsible for this acceleration is due to electric field between the plates.
Electric force due to electric field E is given by F = E x q
Acceleration is given by
a = force/mass
∴ a = force/mass
∴ a = force/mass
∴ a = F/m
∴ a = Eq/m …… (1)
for vertical motion velocity in the vertical direction is zero u =0
using law of motion.
S = ut + at2 …….(2)
Where t is the time taken by particle to cross the field of length L and given by
t = L/vx
From equation 1 and 2 we get,
This is the deflection in vertical direction.
1.34)
In case you are missed :- NCERT Solution for Current Electricity
For more update, follow this page ⇒ NCERT Solutions