Physics Ncert_exemplar 2019 Solutions for Class 12 Science Physics Chapter 1 Electric Charges And Fields are provided here with simple step-by-step explanations. These solutions for Electric Charges And Fields are extremely popular among Class 12 Science students for Physics Electric Charges And Fields Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Physics Ncert_exemplar 2019 Book of Class 12 Science Physics Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Physics Ncert_exemplar 2019 Solutions. All Physics Ncert_exemplar 2019 Solutions for class Class 12 Science Physics are prepared by experts and are 100% accurate.
Page No 1:
Question 1:
(a) shall increase along the positive x-axis.
Answer:
q1 is a negative charge as the net force acting on it is along the positive x-axis.
When the Positive Charge Q is added at the point shown in the figure the force of attraction on the charge q1 will increase and it will get attracted more towards the charge Q or we can say that the force on it shall increase along the positive x-axis.
Hence, the correct answer is option (a).
Page No 2:
Question 2:
A point positive charge is brought near an isolated conducting sphere. The electric field is best given by
Answer:
Positive charge acts as the source of electric field lines. So the field lines will originate from the charge q.
Also, the field lines are always perpendicular to the surface of the conductor.
Tangent to the field line at any point gives the direction of the field at that point.
Due to the presence of the positive charge q near the conducting sphere, negative charge will be induced on the left surface of the sphere and positive charge on the right surface of the sphere.
Only figure 1 follows all of the above points.
Hence, the correct answer is option (a).
Page No 2:
Question 3:
Answer:
According to the Gauss's Law electric flux out of a closed surface is equal to the net charge enclosed within the gaussian surface divided by the permittivity,
Since the charge enclosed by each of the figures is equal to q so the net electric flux through each sphere will be the same.
Hence, the correct answer is option (d).
Page No 3:
Question 4:
Which of the following statements is correct?
Answer:
According to the gauss law, the electric flux coming out of a closed surface depends only on the charge enclosed within that closed surface.
The electric field also depends on the charge enclosed within the closed gaussian surface.
The net charge enclosed within the gaussian surface is q2 & q4 only.
Hence, the correct answer is option (b).
Page No 3:
Question 5:
Answer:
It can be seen from the figure that the field lines are not uniformly spaced, So it can be concluded that the field is not uniform.
Electric field strength decreases from left to right as the crowding of the field lines decreases, So the resultant force on charges also decreases from left to right.
Force on a charge -q is greater than the force exerted by the field on the charge +q.
The dipole will experience a net force towards the left.
Hence, the correct answer is option (c).
Page No 3:
Question 6:
Answer:
When a small point charge is placed near a plane sheet of charge negative charge is induced on that side of the sheet which faces the +q charge whereas the charge induced on the other side of the sheet is opposite in nature i.e. +q
Hence, the correct answer is option is option (a).
Page No 3:
Question 7:
Answer:
The component of the electric field parallel to the diameter cancels out and we are only left with the perpendicular components.
The resultant electric field is perpendicular to the diameter.
Hence, the correct answer is option (a).
Page No 4:
Question 8:
Answer:
The number of field lines entering the surface should be equal to the number of field lines leaving the surface So that the net flux becomes equal to zero.
According to the gauss law, the following condition should be satisfied
The charge enclosed within the gaussian surface is zero.
Hence, the correct answers are option (c) & (d).
Page No 4:
Question 9:
Answer:
The electric field in general is continuous but the field at a point cannot be continuous always.
The electric field at a point is continuous if there is no charge at that point.
It will be discontinuous if there is a charge at that point.
Hence, the correct answer is option (b) and option (d).
Page No 4:
Question 10:
(a) on my surface.
(b) if the charge is outside the surface.
(c) could not be defined.
(d) if charges of magnitude q were inside the surface.
Answer:
Gauss law states that the electric flux through a closed surface depends on the charge enclosed within that closed surface.
So, If there is no charge inside the closed surface then the flux will be equal to zero.
The flux will be non-zero if there is a charge inside the Gaussian surface.
Hence, the correct answer is option (b) and (d).
Page No 4:
Question 11:
Answer:
In the region the net charge is zero, So it can be said that inside the region there is a presence of equal and opposite charges inside the region which is basically a dipole.
So it can be said that the dipole is present inside the region and now we need to calculate the Electric Field at any Point outside this charge configuration.
The electric field due to the dipole at a large distance varies as
Since the electric field is a conservative field so the work done in moving a charged particle in closed loop in the Electric field region is
zero.
So, the correct answers are option (c) and (d).
Page No 4:
Question 12:
(a) total flux through the surface of the sphere is
(b) field on the surface of the sphere is
Answer:
The flux through the closed gaussian surface is given by the Gauss law:
The net charge enclosed within the gaussian surface is given by
So the electric flux through the sphere is
5Q charge lies outside the sphere, So the electric flux due to the 5Q charge is zero.
Hence, The correct answer is option (a) and (c).
Page No 5:
Question 13:
(a) If q > 0 and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre.
Answer:
At the center of the ring, the net electric field is zero. So the net force experienced by the charge q is zero. If the charge is displaced from the center of the ring then there will be an electric field in the opposite direction. So the charge will be pushed back to its original position.
A similar thing will happen for -q charge and it will also be pushed back to its original position by the above criteria.
The charge -q will perform SHM for small displacement.
The point at the center is a point of unstable equilibrium.
Page No 5:
Question 14:
Answer:
The net charge enclosed within the closed surface is given by the sum of the two charges of the dipole i.e. -q+q =0
So the net electric flux is given by the formula
So, the net electric flux through this closed surface is zero.
Page No 5:
Question 15:
Answer:
The charge induced on the inner surface is -Q
The charge on the outer surface of the sphere is +Q.
The surface charge density is given by charge divided by the Area of the surface.
Page No 5:
Question 16:
The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero?
Answer:
There is an equal number of protons and neutrons present in a neutral atom, So the net charge on an atom is zero.
Page No 5:
Question 17:
Answer:
According to the gauss law, the electric field through a closed Gaussian surface depends only on the charge enclosed within that gaussian surface.
Mathematically,
But the Electric field at the surface could also be due to a charge present outside the closed gaussian surface.
So, the net charge inside the closed gaussian surface decides the field coming out of the gaussian surface only, It does tell us about the field due to any external charges, which could be present.
So it would be wrong to say that the electric field at the surface is zero is the net charge enclosed within the surface is zero.
Now if the Electric field on the surface is zero, then the charge enclosed within the gaussian surface will necessarily be zero. This is the correct statement.
Page No 5:
Question 18:
Sketch the electric field lines for a uniformly charged hollow cylinder shown in the given figure.
Answer:
ans
Page No 6:
Question 19:
Answer:
(a) Using the symmetry we can draw the Gaussian surface such that it encloses the corner A.
Total flux through one of the cube is given by
(b) Again we can draw the Gaussian surface such that it enclose the charge completely. So the flix is now given by
(c) The flux, in this case, is given by
(d) The flux, in this case, is also given by
Page No 6:
Question 20:
Answer:
Mass of paisa coin = 0.75 g (made up Al only)
Atomic mass of Al = 27u
Number of moles of 0.75 g of
Number of Aluminium atoms = number of moles × NA
Where, NA is the Avogadro number.
Thus, number of Al atom
But, atomic number of Al = 13
That means the atom of Aluminium has 13e– and 13p+ in neutral state.
So, number of either electron and proton:
Total no. of e– = no. of atoms × no. of e– in one atom
Total Charge = 3.48 × 104 C
The coin has equal amount of positive and negative charge 3.48 × 104C.
Page No 6:
Question 21:
Answer:
Given:
q1 = 3.48 × 104 C
q2 = –3.48 × 104 C
(i) r = 1 cm = 10–2 m
(ii)Similarly for, r = 100 m = 102 m
(iii) Similarly for, r = 106 m
Page No 6:
Question 22:
Answer:
(i) Due to symmetry of figure, some of all attractive force acting on the chlorine atom is equal to zero.
Thus,
Hence, the net electric field will be equal to zero.
(ii) If the Cs atom at 'A' is missing then force on Cl due to other seven:
Page No 6:
Question 23:
Answer:
Let the charge is placed at point 'P', at a distance 'x' on left of charge 'q' as shown below,
At P, the charge '2q' experience force towards left due to 'q' and towards right due to '–3q'
Solving the quadratic equation we get:
Negative sign would be between 'q' and '–3q' and hence it is not acceptable.
Thus, to left of q.
Page No 7:
Question 24:
Answer:
(a) A and C are positive charges. (Electric field lines are outwards).
(b) The electric field line density is maximum at point C. Hence, the charge 'C' has the largest magnitude.
(c) (i) near A
Page No 7:
Question 25:
(a) (i) What will be the electric field at O, the centre of the pentagon?
Answer:
(a)
(i) Due to symmetry of figure, the electric field at centre 'O' of the polygon is zero.
(ii) When the charge 'q' is removed then net electric field due to the other charges at O,
(along OA)
(iii) If charge 'q' at A is replaced by '–q' then the value of net electric field at O,
(along OA.)
(b) Similar method can be applied as in part (a) to calculate the electric field for regular polygon of n sides (symmetry is used to solve).
Page No 7:
Question 26:
Answer:
(a) Let us suppose that universe is a perfect sphere of radius R.
If 'E' is the electric field intensity at a distance 'R' on the surface of the sphere, then according to Gauss's theorem
Coulomb force acting on one hydrogen atom,
Positive sign shows repulsive force.
Gravitational potential at boundary of universe =
M = mass of universe (All H atoms)
So, the gravitational force acting on H atom at boundary universe = Gravitational potential × mH
Where, mH = Mass of hydrogen atom = Mass of one proton (mp)
∴ M = Mass of universe = Number of H atoms ×mp
If FC > FG then the universe will start to expand.
For critical value of expansion, FC = FG
Solving we get,
So critical angle of 'y' is of the order of 10–18 so that universe starts to expand.
(b) Net force on hydrogen atom is given by,
Solving we get, R = Aeαt + Be–αt where, A and B are constants.
We are looking for expansion,
So, B = 0 and R = Aeαt
⇒ Velocity of expansion,
.
.
Hence, v R i.e., velocity of expansion is proportional to distance from the centre.
Page No 8:
Question 27:
Answer:
(a) Let us consider a sphere 'S' of radius 'R' and two hypothetical sphere of radius r < R and r > R.
Here, charge density is positive.
So, direction of E is radially outwards.
For point r > R, electric field intensity is given by,
Direction of E is radically outwards.
(b) The two protons must be on the opposite sides of the centre along a diameter following the rule of symmetry:
If proton 1 and 2 are embedded at distance 'r' from the centre of the sphere as shown, then attractive force on proton 1 due to charge distribution is
Repulsive force on proton 1 due to proton 2 is,
Net force on 1,
F = F1 + F2
So,
Thus, net force on proton 1 will be zero, when
Page No 8:
Question 28:
Answer:
(a) Net electric field at plate γ before collision is equal to the sum of electric field at plate γ due to plate α and β.
The electric field at plate γ due to plate β is
Hence, the net electric field at plate γ before collision.
to the left, if Q > q.
(b) During collision, plate β and γ are together. Their potential becomes same.
At any point O, in between the two plates, the electric field must be zero.
As the electric field at O is zero, therefore
Q + q2 = q1
⇒ Q = q1 − q2 .....(1)
As there is no loss of charge in collision,
Q + q = q1 + q2 .....(2)
On solving equation (1) and (2)
(c) After collision, at a distance 'd' from β,
After the collision, electric field at plate γ is
Just before collision, electric field at plate γ is
If F1 is force on plate γ before collision, then
Total work done, ω = (F1 + F2) d,
If 'm' is mass of plate 'γ', the K.E. gained by plate
According to work energy principle,
Page No 8:
Question 29:
Answer:
(i)
So, 1 esu of charge = × 1 cm
⇒ 1 esu of charge
Thus, esu of charge is represented in terms of fractional powers of M and of L.
(ii) Let 1 esu of charge = x C, where 'x' is a dimension less number.
This situation is equivalent to two charges of magnitude x C separated by 10–2 m.
Taking,
We get,
Page No 9:
Question 30:
Answer:
Given, two charge '–q' at 'A' and 'B'.
AB = AO + OB = 2d
x = small distance perpendicular to O.
So, force of attraction at 'P' towards A and B are each
where AP = BP = r.
Horizontal components of these forces Fn are cancel out. Vertical components along PO add.
If ∠APO = 0, the net force on 'q' along PO is F′
⇒ F′ = 2F cosθ
When,
Where,
⇒ F ∝ x
Force on charge q is proportional to its displacement from the centre O and it is directed towards O.
Hence, motion of charge 'q' would be the simple harmonic, where
and
Page No 9:
Question 31:
(a) Show that the particle executes a simple harmonic oscillation.
(b) Obtain its time period.
Answer:
(a) Let the charge 'q' is displaced towards the axis of ring.
Let,
So, Fnet = KZ
Force is directly to the Z and directed towards O.
So, net force can be written as:
Fnet = –KZ.
This shows particles executes SHM.
(b)
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