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1.	The electrostatic force on a charge due to another charge is F. In presence of one more charge, what would be the electrostatic force on first charge due to second charge?
Answer» In the presence of other charge there will be no effect on electrostatic force. Thus electrostatic force will remain same (F).

2.	Explain the quantization of charge.
Answer» Charge quantization means that charge cannot take any arbitrary values but only values that are integral multiples of the fundamental charge (charge of electron/proton). The basic cause of quantization is that only integral no. of electrons can be transferred from one body to another. Thus an object’s charge can be exactly 0 or exactly q = ±ne i.e., 1e, -1e, 2e, -2e…. etc.

3.	Write the superposition principle for electrostatic forces.
Answer» The principle of superposition states that when a number of charges are interacting, the net electrostatic force an a given charge is the vector sum of the forces exerted on it due to all other charges. The force between two charges is not affected by the presence of other charges.

4.	In an electric field E, a unit negatively charged ion and an electron are moving. Which one of them move fast and why?
Answer» Electron will move with fast speed because its mass is less than unit negative charged ion. Due to capture of electron, the mass of unit negative charged ion will be more.

5.	The electric field intensity at a point on the line joining two point charges is zero. What can you conclude about the charges?
Answer» It means both the charges are of same nature, so the electric field intensity is zero.

6.	Electric field decreases when going far from the point charge. This is true for an electric dipole. Does the electric field decreases at the same rate in both the cases?
Answer» For electric dipole at axial and equatorial positions, for far points (r >> 2l) electric field E ∝ \(\frac{1}{r^3}\) and for single charge, electric field E ∝ \(\frac{1}{r^2}\) case of electric dipole electric field decreases comparatively for single charge.

7.	Write the value of one quantum charge.
Answer» One quantum charge = 1.6 × 10^-19C

8.

If a glass plate is placed between two charges. Then, compare the electric force with the initial condition. It will be:(a) more(b) less(c) zero(d) infinite

Answer»

(b) less

On placing the glass plate between the charge, the electric field will decrease as comparison to previous.

9.	What is meant by frictional electricity? Give the brief description of its origin.
Answer» Frictional electricity is the electricity produced by rubbing two suitable bodies and transfer of electrons from one body to other. The body which looses the electrons, becomes positively charged while the body which receives the electrons, becomes negatively charged.

10.	The dipole moment of HCl molecule is 3.4 × 10-30 cm. The distance between its ions will be:(a) 2.12 × 10-11 m(b) zero(c) 2 mm(d) 2 cm
Answer» (b) zero Ions are bound to each other, thus the distance between them will be zero.

11.	What is the resultant force on an electric dipole placed in uniform electric field?
Answer» The resultant force on an electric dipole placed in uniform electric field is Zero.

12.	An electron and a proton are at a distance of 1 A. The dipole moment of the system is:(a) 3.2 × 10-29 Cm(b) 1.6 × 10-19 Cm(c) 1.6 × 10-29 Cm(d) 3.2 × 10-19 Cm
Answer» (c) 1.3 × 10^-29Cm p = q × 2l = 1.6 × 10^-19 × (1× 10^-10) = 1.6 × 10^-29Cm

13.	An electric dipole is placed in a uniform electric field. Show that it will not accelerate.
Answer» In uniform electric field on electric dipole two equal forces will act in opposite direction, which produce a torque. Therefore electric dipole will not perform translatory accelerating motion.

14.	What is the condition of an ideal electric dipole?
Answer» We can think of an ideal dipole in which size 2α → 0 and charge q → ∞ in such a way that the dipole moment, p = q × 2α a has a finite value, such a dipole of negligibly small size is called an ideal or point dipole.

15.	Give an example of a particle for which the rest mass is zero and uncharged.
Answer» The particle is photon.

16.	On what factors does the constant k = \(\frac{1}{4 \pi \varepsilon_{0}}\) depends?
Answer» Constant k = \(\frac{1}{4 \pi \varepsilon_{0}}\) depends on the medium and system of measurement.

17.	A copper sphere of 2 gm has 2 × 1022 atoms. The charge on the nucleus of each atom is 29e. How many fraction of electrons be eliminated to provide a charge of 2μC to the sphere?
Answer» Charge on each nucleus = 22 eC ∴ Net charge on nucleus of sphere of mass 2 gm = (22 e) × (2 × 10²²C) = (4.4 × 10²³)eC ∴ Total number of electron on sphere = (4.4 × 10²³) [∵ Sphere is initially uncharged] ∴ Number of electrons in charge of 2 × 10^-6)C = \(\frac{2 \times 10^{-6}}{1.6 \times 10^{-19}}\) = 1.25 × 10¹³ ∴ Fraction of electrons, which is to be removed = \(\frac{1.25 \times 10^{13}}{4.4 \times 10^{23}}\) = 2.84 × (10²)^-11

18.

Write the value of charge (in coloumbs) on the nucleus of 7N14.

Answer»

As q = Ze

q = 7e = 7 × 1.6 × 10^-19C

= 11.2 × 10^-19C