Repulsive force of magnitude 6 x 10⁻³ N
Charge on the first sphere, q₁ = 2 x 10⁻⁷ C
Charge on the second sphere, q₂ = 3 x 10⁻⁷ C
Distance between the spheres, r = 30 cm = 0.3 m
Electrostatic force between the spheres is given by the relation,
\(F=\frac{q_{1} q_{2}}{4 \pi \in_{0} r^{2}}\)
Where, ∈₀ = Permittivity of free space
\(\frac{1}{4 \pi \in_{0}}=9 \times 10^{9} N m^{2} C^{-2}\)
\(F=\frac{9 \times 10^{9} \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^{2}}=6 \times 10^{-3} N\)
Hence, force between the two small charged spheres is 6 x 10⁻³ N. The charges are of same nature. Hence, force between them will be repulsive.

In one second 10⁹ electrons move out of the body. Therefore the charge given out in one second is 1.6 x 10^–19 × 10⁹ C = 1.6 x 10^–10 C. The time required to accumulate a charge of 1 C can then be estimated to be 1 C ÷ (1.6 x 10^-10 C/s) = 6.25 x 10⁹ s = 6.25 x 10⁹ ÷ (365 x 24 x 3600) years = 198 years. Thus to collect a charge of one coulomb, from a body from which 10⁹ electrons move out every second, we will need approximately 200 years. One coulomb is, therefore, a very large unit for many practical purposes.
It is, however, also important to know what is roughly the number of electrons contained in a piece of one cubic centimetre of a material. A cubic piece of copper of side 1 cm contains about 2.5 x 10²⁴ electrons.

When a glass rod is rubbed with a silk cloth, charges appear on both. These charges are equal in magnitude and opposite in sign, so that algebraic sum of the charges produced on both is zero. The net charge on the two bodies was zero even before rubbing them. Thus, we find that charges can be created only in equal and unlike pairs. This is consistent with the law of conservation of charge.

(a) Electric charge of a body is quantized. This means that only integral (1, 2, …., n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge
(b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.

Let us assume that the mass of one cup of water is 250 g. The molecular mass of water is 18g. Thus, one mole (= 6.02 x 10²³ molecules) of water is 18 g. Therefore the number of molecules in one cup of water is (250/18) x 6.02 x 10²³.
Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. Hence the total positive and total negative charge has the same magnitude. It is equal to (250/18) x 6.02 x 10²³ x 10 x 1.6 x 10^–19 C = 1.34 x 10⁷ C.

(a) (i) The electric force between an electron and a proton at a distance r apart is:
\(F_{e}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}\)
where the negative sign indicates that the force is attractive. The corresponding gravitational force (always attractive) is:
\(F_{G}=-G \frac{m_{p} m_{e}}{r^{2}}\)
where m_p and m_e are the masses of a proton and an electron respectively
\(\left|\frac{F_{e}}{F_{G}}\right|=\frac{e^{2}}{4 \pi \varepsilon_{0} G m_{p} m_{e}}=2.4 \times 10^{39}\)
(ii) On similar lines, the ratio of the magnitudes of electric force to the gravitational force between two protons at a distance r apart is :
\(\left|\frac{F_{e}}{F_{G}}\right|=\frac{e^{2}}{4 \pi \varepsilon_{0} G m_{p} m_{p}}=1.3 \times 10^{36}\)
However, it may be mentioned here that the signs of the two forces are different. For two protons, the gravitational force is attractive in nature and the Coulomb force is repulsive . The actual values of these forces between two protons inside a nucleus (distance between two protons is ~ 10^-15 m inside a nucleus) are F_e ~ 230 N whereas F_G ~ 1.9 x 10^–34 N.
The (dimensionless) ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces.
(b) The electric force F exerted by a proton on an electron is same in magnitude to the force exerted by an electron on a proton; however the masses of an electron and a proton are different. Thus, the magnitude of force is
\(|\mathbf{F}|=\frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}=8.987 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2} \times\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2} /\left(10^{-10} \mathrm{~m}\right)^{2}\)
= 2.3 × 10^–8 N
Using Newton’s second law of motion, F = ma, the acceleration that an electron will undergo is
a = 2.3 x 10^–8 N / 9.11 x 10^–31 kg = 2.5 x 10²² m/s²
Comparing this with the value of acceleration due to gravity, we can conclude that the effect of gravitational field is negligible on the motion of electron and it undergoes very large accelerations under the action of Coulomb force due to a proton.
The value for acceleration of the proton is
2.3 x 10^–8 N / 1.67 x 10^–27 kg = 1.4 x 10¹⁹ m/s².

(i) (d): Electric flux \(\phi=\vec{E} \cdot \vec{A}=E A \cos \theta\)
where \(\vec{A}=A \hat{n}\)
For electric flux passing through \(S_{6}, \hat{n}_{S_{6}}=-\hat{i} \text { (Back) }\)
\(\therefore \quad \phi_{S_{6}}=-\left(4 \times 10^{3} \mathrm{NC}^{-1}\right)(0.1 \mathrm{~m})^{2} \cos 37^{\circ}\)
= -32 N m² C-I
(ii) (a) : For electric flux passing through S_1,
\(\hat{n}_{S_{1}}=-\hat{j} \text { (Left) }\)
\(\therefore \quad \phi_{S_{1}}=-\left(4 \times 10^{3} \mathrm{~N} \mathrm{C}^{-1}\right)(0.1 \mathrm{~m})^{2} \cos \left(90^{\circ}-37^{\circ}\right)\)
= -24 N m² C^-1
(iii) (c): Here, \(\hat{n}_{S_{2}}=+\hat{k}(\text { Top })\)
\(\therefore \quad \phi_{S_{2}}=-\left(4 \times 10^{3} \mathrm{~N} \mathrm{C}^{-1}\right)(0.1 \mathrm{~m})^{2} \cos 90^{\circ}=0\)
\(\hat{n}_{S_{3}}=+\hat{j} \text { (Right) }\)
\(\hat{n}_{S_{4}}=-\hat{k}(\text { Bottom })\)
\(\therefore \quad \phi_{S_{4}}=\left(4 \times 10^{3} \mathrm{~N} \mathrm{C}^{-1}\right)(0.1 \mathrm{~m})^{2} \cos 90^{\circ}=0\)
And, \(\tilde{n}_{S_{5}}=+\hat{i} \text { (Front) }\)
\(\therefore \quad \phi_{S_{5}}=+\left(4 \times 10^{3} \mathrm{~N} \mathrm{C}^{-1}\right)(0.1 \mathrm{~m})^{2} \cos 37^{\circ}\)
= 32 N m² C^-1
S₂ and S₄ surface have zero flux.
(iv) (d): As the field is uniform, the total flux through the cube must be zero, i.e., any flux entering the cube must leave it.
(v) (b): Surface integral \(\oint \vec{E} \cdot d \vec{S}\)is the net electric flux over a closed surface S.
\(\therefore \quad\left[\phi_{E}\right]=\left[\mathrm{ML}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]\)

(i) (b): The proportionality constant k depends on the nature of the medium between the two charges.
(ii) (c): \({ As, }\left[\varepsilon_{0}\right]=\frac{1}{4 \pi F} \cdot \frac{q_{1} q_{2}}{r^{2}} =\frac{[\mathrm{AT}]^{2}}{\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]\left[\mathrm{L}^{2}\right]} \)
\(=\left\lfloor\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]\)
(iii) (b)
(iv) (d): \(F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{d^{2}}\)
\(\therefore\left(10 \times 10^{-3}\right) \times 10=\frac{\left(9 \times 10^{9}\right) \times q^{2}}{(0.6)^{2}}\)
\(\text { or } \ q^{2}=\frac{10^{-1} \times 0.36}{9 \times 10^{9}}=4 \times 10^{-12}\)
\(\text { or } \ q=2 \times 10^{-6} \mathrm{C}=2 \mu \mathrm{C}\)
(v) (b)