The given ratio is
\(\frac{k e^{z}}{G m_{c} m_{p}}\)
Where,
G = Gravitational constant
Its unit is N m² kg⁻².
m_e and m_p = Masses of electron and proton.
Their unit is kg.
e = Electric charge.
Its unit is C.
k = A constant
\(=\frac{1}{4 \pi \in_{0}}\)
∈₀ = Permittivity of free space
Its unit is N m² C⁻².
therefore, unit of the given ratio \(\frac{k e^{2}}{G m_{c} m_{p}}=\frac{\left[N m^{2} C^{-2}\right]\left[c^{-2}\right]}{\left[N m^{2} k g^{-2}\right][k g][k g]}=m^{0} L^{0} T^{0}\)
Hence, the given ratio is dimensionless.
e = 1.6 x 10⁻¹⁹ C
G = 6.67 x 10⁻¹¹ N m² kg^-2
m_e= 9.1 x 10⁻³¹ kg
m_p = 1.66 x 10⁻²⁷ kg
Hence, the numerical value of the given ratio is
\(\frac{k e^{2}}{G m_{c} m_{p}}=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{6.67 \times 10^{-1} \times 9.1 \times 10^{-3} \times 1.67 \times 10^{-22}} \approx 2.3 \times 10^{39}\)
This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

(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.

(a) Yes, by negligible amount.
(b) Yes.
(c) The charging of comb is due to charging by friction. The charging of paper bits is due to charging by induction.

For like charges, \(\vec{F}_{12}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{3}} \vec{r} 21\)
where \(\vec{F}_{12}\) is the force experienced by the 1st charge due to 2nd one and \(\overrightarrow{r_{21}}=\vec{r}_{1}-\vec{r}_{2}\)

Similarly, \(\overrightarrow{F_{21}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{3}} \overrightarrow{r_{12}}, \overrightarrow{r_{12}}=\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\)
\(\because \ \overrightarrow{r_{12}}=-\overrightarrow{r_{21}}\)
\(\overrightarrow{F_{12}}=-\overrightarrow{F_{21}}\)

∴ Newton's third law is obeyed by the Coulomb's law.

If the field is non-uniform, the net force will be non-zero.

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.

Electrons have been removed from the object.

We know that, q = ne 
\(\Rightarrow 1=\mathrm{n} \times 1.6 \times 10^{-19}\)   (given, q = 1C)
\(\text { i.e. } n=\frac{1}{1.6 \times 10^{-10}} \simeq 6 \times 10^{18}\)
∴ 1 C is the charge of 6 x 10¹⁸ electrons.
So, a coulomb of charge is bigger than the charge on an electron.

The basic cause of quantisation of charge is only the integral number of electrons which is transferred from one body to another, i.e. ± ne.

The charges q₁ and q₂ are equal and opposite.

It is also attractive in nature.

(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)