12th physics chapter 1 subjective questions in english

1. Define charge.

Answer:

It is defined as the basic and characteristic property of elementary particles of matter in the form of which certain force of interaction and interaction energies may be explained.

2. Is mass of A body affected on charging?

Answer:

Yes, it is affected on charging.

3. By how much amount the mass of a body is affected on charging?

Answer:

Mass will increase or decrease equal to the mass of electrons gained or lost.

4. Write the relation between 1 Coulomb and 1 e.s.u. of charge.

Answer:

1 C = 3 x 10⁹ e.s.u.

5. Is electric field intensity a scalar or a vector quantity?

Answer:

Electric field intensity is a vector quantity.

6. Write down the S.I. unit of electric field intenstiy.

Answer:

Newton (Coulomb^-1) = NC^-1

14. Tell the nature of Torque.

Answer:

Torque is a vector quantity.

7. What is the direction of electric field due to an electric dipole at a point on its equitorial line?

Answer:

The electric field due to an electric dipole at a point on its equitorial line acts opposite to the direction of dipole moment of the electric dipole.

8. What is the direction of electric field due to an electric dipole at a point on its axial line?

Answer:

The electric field due to an electric dipole at a point on its axial line acts in the direction of dipole moment of the electric dipole.

9. What is the nature of symmetry of field due to an electric dipole?

Answer:

It is of cylindrical nature.

10. What can be concluded if the electric flux through a closed surface is zero?

Answer:

It can be concluded from it that the inward electric flux is equal to the outward electric flux.

11. Define a Gaussian surface.

Answer:

It is defined as an imaginary closed surface enclosing some Charge inside it.

12. If electric flux through a closed surface is negative, then what type of charge it contains?

Answer:

It contains negative charge.

13. What is the importance of Gauss’s theorem?

Answer:

Gauss’s theorem helps us to calculate the electric field intensity ‘ in those cases where it becomes quite difficult to calculate it using Coulomb’s law.

14. A Gaussian surface contains charges of – q, + 2q and- q. Calculate the electric flux through the surface.

Answer:

According to Gauss’s law, we know that the electric flux is given by

ϕ=q/ε₀ …………(1)
Here q = total charge enclosed bv the Gaussian surface = – q + 2q + (-q) = 0.
∴ From (1) and (2), we get
ϕ = 0/ε₀ = 0
Φ = 0

15. Does Coulomb’s law and Gauss’s law complement each other? How?

Answer:

Yes, because Coulomb’s law can be derived from Gauss’s law and vice-versa.

16. Define friction electricity. Why it is called static electricity?

Answer:

It is defined as the electricity produced on the objects when they are rubbed with each other. It is called static electricity because the electric charges so developed cannot move from one part of the object to its other part.

17. Does, the electric field due to an infinite plane sheet of charge depends upon the distance of the point from the sheet?

Answer:

No, it does not depend upon the distance of the point from the sheet.

18. Why does a min inside an insulated metallic cage not ,. receive a shock when the cage is highly charge.

Answer:

It is because, the potential at each point inside the cage is same as that of the cage itself. Since there is no potential difference between the man and the highly charged cage, so the man does not receive any shock.

19. A positively charged glass rod attracts a suspended pith ball. Does it mean that the pith-ball is negatively charged?

Answer:

No. The pith-ball can be uncharged also. The positively charged glass rod can attract the pith-ball due to induced charges of opposite kind produced on the pith-ball.

20. Is Coulomb’s law in electrostatics applicable in all situations?

Answer:

No, Coulomb’s law in electrostatics is not applicable in all situations.

21. Tell the situations in which Coulomb’s law is applicable.

Answer:

It is applicable in the following situations :
(i) The electric charges must be stationary.
(ii) The electric charges must be point charges in sizes.

22. What does q₁+ q₂= 0 signify in electrostatic.

Answer:

It signifies that the two charges q₁ and q₂ are equal and opposite and constitute an electric dipole.

23. What are the properties of electric charges?

Answer:

The following are the properties of electric charges.

They are always additive in nature.

The charge is a scalar quantity.

The charge is of two types i.e., + ve and – ve.

The charge is always quantised.

The charge of an isolated system always remains conserved.

The charge on an object unlike mass is not affected by the motion of the object.

Similar charges repel each other and unlike.charges attract each other.

24. Can test charge be equal to zero? Why?

Answer:

No, the test charge q₀ cannot be zero because it will not experience any force and thus there would be no way to measure electric field.

25. Give important porperties of electric lines of force?

Answer:

The following are the properties of electric lines of force:

They start from positive charge and terminate on the negative charge.

They don’t pass through a conductor.

The lines of force never intersect each other.

The lines of force always contract longitudinally. This explains the attraction between two dissimilar charges.

The lines of force exert a lateral pressure on each other which explains the repulsion between two similar charges.

The tangent drawn at any point to the electric lines of force gives the direction of electric field at that point.

The lines of force are closer to each other where the electric field is stronger and spread farther apart where the electric field is weaker.

26. A boy brings the palm of his hand near the disc of a charged gold leaf electroscope. The leave of the electroscope are observed to collapse slightly. But when the boy moves his hand away from it, the leaves resume their original position. How can you explain this behaviour of leaves?

Answer:

While an equal and opposite charge will be induced on the lower side of the palm, equal and same charge will be induced on the upper side of the palm. But this will leak to the earth. The induced opposite charge causes the leaves to get slightly collapsed. ‘

27. A metal sphere is held fixed on a smooth horizontal insulated plate and another metal sphere is placed some distance away. If the fixed sphere is given a charge, how will the other sphere react?

Answer:

When the fixed sphere is given a charge, it induces opposite charge on the nearer end of the other sphere and similar charge on the farther end. The net force is attractive in nature. So, the free sphere will be accelerated towards the fixed sphere.

28. Mention two similarities and two dissimilarities between the electrostatic and gravitational forces.

Answer:

Similarities
(i) Both the electrostatic and gravitational forces are conservative in nature.
(ii) Both of them follow inverse square law i.e., these forces are inversely proportional to the square of the distance between the two charges or two masses.

Dissimilarities –

(i) Electrostatic forces between two point charges can be attractive or repulsive while the gravitational forces are only attractive in nature. ‘
(ii) Electrostatic forces depend on the medium in which the two charges are placed while gravitational force does not depend on the mediumbetween the two bodies.

29. (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:

(a) The meaning of the statement ‘electric charge of a body is quantised’ is that the charge on it is always some integral multiple of elementary charge of an electron or a proton (= e in magnitude) i.e., charge on a body never varies continuously but it varies in the form of discrete packets called quanta or packets of charge. Mathematically, the charge on a body can be expressed as
q = ±ne
where n is an integer, e = magnitude of the charge of an electron or proton = 1.6 x 10^-19 C. A fraction of the fundamental charge e has never been observed in free state.

(b) In practice, the charge on a charged body at macroscopic level is very large while the charge on an electron is very small. When electrons are added to or removed from a body, the change taking place in the total charge on the body is so small that the charge seems to be varying in a continuous manner. Thus quantisation of electric charge can be ignored at macroscopic level i.e., when dealing with a large scale charged body.

30. 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 conservaton of charge.

Answer:

Initially i.e., before rubbing both the glass rod and silk cloth are electrically neutral. In other words, net charge On the glass rod and silk cloth is zero. When the glass rod is rubbed with silk cloth, a few electrons get transferred from the rod to the silk cloth, thus glass rod becomes positively charged and silk cloth negatively charged. The positive charge on the glass rod is exactly equal to the negative charge on the silk cloth, so net charge on the system is again zero.

Thus the appearance of charge on the glass rod and silk cloth is in accordance with the law of conservation of charge as the total charge of the isolated system is constant. Similarly when ebonite rod is rubbed with fur, they acquire – ve and + ve charges respectively and net charge is zero again. Thus we conclude that charge is neither created nor destroyed but it is merely transferred from one body to another which is consistent with the law of conservation of charge.

31. (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

(b) Explain why two fields lines never cross each other at any point?

Answer:

(a) The electrostatic line of force is the path tangent at every point of which gives the direction of electric field at that point. The direction of electric field generally changes from point to point. So the lines of force are generally curved lines. Further they are continuous curves and cannot have sudden breaks because if it is so, then it will indicate the absence of electric field at the break points.

(b) The electric lines of force never cross each other because if they do so, then at the point of their intersection, we can draw two tangents which give two directions of electric field at that point which is not possible.

32. An electric dipole with dipole moment 4 x 10-9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 x 10-4 NC

Calculate the magnitude of the torque acting on the dipole.

Answer:
Here, p = 4 x 10^-9 Cm, E = 5 x 10⁴NC^-1,θ = 30°, τ = ?
Using the formula, x = pE sin θ,

we get
τ = 4 x 10^-9 x 5 x 10⁴ x sin 30°
= 20 x 10^-5 x 12
= 10^-4 Nm.

33. A polythene piece rubbed with wool is found to have a negative charge of 3 x 10^-7

(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 polvihene piece.
From quantisa Lion of charge, we know that q = ne.
n = qe = −3×10⁷−1.6×10⁻¹⁹ = 1.875 x 10¹²
= 2 x 10¹²

(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^-13 kg,
n = 1.875 x 10¹²
M = total mass transferred to polythene = ?
= m x n
= 9.1 x 10³¹ x 1.875 x 10¹² = 1.71 x 10^-18kg ≈ 2 x 10^-18 kg.

34. 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 becuase the number of lines entering the cube of side 20 cm is same as the number of lines leaving the cube.

Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 x 10³ Nm²/c

(i) What is the net charge inside the box?
(ii) If the net outward flux through the surface of the box were zero,

35. could you conclude that there were no charges inside the box? Why or why not?

Answer:

(i) Given, Φ = 8 x 10³Nm² C^-1, ε₀= 8.854 x 10^-12 C² N^-1m^-2

If the net charge inside the black box is q, then using formula
Φ = qε₀ or we get, q = E₀ Φ
or q = 8.854 x 10^-12 x 8 x 10³C
q = 8.854 x 8 x 10^-9 C
q = 70.832 x 10^-9 C = 0.070832 x 10^-6C = 0.071 pC

(ii) We cannot conclude that the net electric charge inside the box
is zero if the outward flux through the surface of black box is zero because there might be equal amounts of positive and negative charges cancelling each other and thus making the resultant charge equal to zero. Thus, we can only conclude that the net charge inside the box is zero.

36. A point charge causes an electric flux of – 1.0 x 10³Nm²/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge, (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?
    Answer:
    Here, = electric flux through the spherical Gaussian surface
    = -1.0 x 10³N m²C^-1.
    r = radius of Gaussian spherical surfaces – 10 cm
    Let q = charge enclosed at its centre.

(a) According to Gauss’s law, the electric flux through a Gaussian surface depends upon the charge enclosed inside the surface and not upon its size. Thus the electric flux will remain unchanged i.e., – 1.0 x 10³ Nm² C^-1 through the spherical Gaussian surface of double radius i.e. of 20 cm as it also encloses the same amount of charge.

(b) q = point charge = ?, ε₀ = 8.854 x 10^-12 Nm^-2 C².
Using the formula,
Φ = qε₀ we get
q = ε₀ Φ = 8.854 x 10^-12 x (-1.0 x 10³) = – 8.854 x 10^-9C = 8.8 nC.