Coulomb's law states that the electrostatic force of attraction or repulsion acting between two stationary point charges is given by
\(F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\)

where F denotes the force between two charges q₁ and q₂ separated by a distance r in free space, E_o is a constant known as permittivity of free space. Free space is vacuum and may be taken to be air practically.
If free space is replaced by a medium, then E_o is replaced by (E_ok) or (E_oEr)where k is known as dielectric constant or relative permittivity.
(i) In coulomb's law, F = \(k \frac{q_{1} q_{2}}{r^{2}}\), then on which of the following factors does the proportionality constant k depends?

(ii) Dimensional formula for the permittivity constant E_o of free space is

(iii) The force of repulsion between two charges of 1 C each, kept 1 m apart in vaccum is

(iv) Two identical charges repel each other with a force equal to 10 mgwt when they are 0.6 m apart in air. (g = 10 ms^-2). The value of each charge is

(v) Coulomb's law for the force between electric charges most closely resembles with

In 1909, Robert Millikan was the first to find the charge of an electron in his now-famous oil-drop experiment. In that experiment, tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops were observed with a magnifying eyepiece, and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force qE just equaled Mg. Millikan accurately measured the charges on many oil drops and found the values to be whole number multiples of 1.6 x 10^-19 C the charge of the electron. For this, he won the Nobel prize.

(i) If a drop of mass 1.08 x 10^-14 kg remains stationary in an electric field of 1.68 x 10⁵ N C^-I, then the charge of this drop is

(ii) Extra electrons on this particular oil drop (given the presently known charge of the electron) are

(iii) A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field 100 V m^-1.If the mass of the drop is 1.6 X 10^-3 g, the number of electrons carried by the drop is (g= 10 m s^-2)

(iv) The important conclusion given by Millikan's experiment about the charge is

(v) If in Millikan's oil drop experiment, charges on drops are found to be \(8 \mu \mathrm{C}, 12 \mu \mathrm{C}, 20 \mu \mathrm{C}\) then quanta of charge is

Gauss's law and Coulomb's law, although expressed in different forms, are equivalent ways of describing the relation between charge and electric field in static conditions. Gauss's law is \(\varepsilon_{0} \phi=q_{\text {encl }}\), when
q_encl is the net charge inside an imaginary closed surface called Gaussian surface. \(\phi=\oint \vec{E} \cdot d \vec{A}\) gives the electric flux through the Gaussian surface. The two equations hold only when the net charge is in vacuum or air.

(I) If there is only one type of charge in the universe, then \((\vec{E} \rightarrow \text { Electric field, } d \vec{s} \rightarrow \text { Area vector })\)

(ii) What is the nature of Gaussian surface involved in Gauss law of electrostatic?

(iii) A charge 10 \(\mu \)C is placed at the centre of a hemisphere of radius R = 10 cm as shown. The electric flux through the hemisphere (in MKS units) is

(iv) The electric flux through a closed surface area S enclosing charge Q is \(\phi\). If the surface area is doubled, then the flux is 

(v) A Gaussian surface encloses a dipole. 'The electric flux through this surface is

In practice, we deal with charges much greater in magnitude than the charge on an electron, so we can ignore the quantum nature of charges and imagine that the charge is spread in a region in a continuous manner. Such a charge distribution is known as continuous charge distribution. There are three types of continuous charge distribution : (i) Line charge distribution (ii) Surface charge distribution (iii) Volume charge distribution as shown in figure.

(I) Statement 1 : Gauss's law can't be used to calculate electric field near an electric dipole.
Statement 2 : Electric dipole don't have symmetrical charge distribution.

(ii) An electric charge of 8.85 X 10^-13 C is placed at the centre of a sphere of radius 1 m. The electric flux through the sphere is

(iii) The electric field within the nucleus is generally observed to be linearly dependent on r. So,

(iv) What charge would be required to electrify a sphere of radius 25 cm so as to get a surface charge density of \(\frac{3}{\pi} \mathrm{C} \mathrm{m}^{-2} ?\)

(v) The SI unit of linear charge density is

Surface charge density is defined as charge per unit surface area of surface charge distribution. i.e., \(\sigma=\frac{d q}{d S}\) Two large. thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs having magnitude of 17.0 x 10^-22 C m^-2 as shown. The intensity of electric field at a point is \(E=\frac{\sigma}{\varepsilon_{0}}\) where \(\varepsilon_{0}\) = permittivity of free space.

(i) E in the outer region of the first plate is

(ii) E in the outer region of the second plate is

(iii) E between the plates is

(iv) The ratio of E from right side of B at distances 2 cm and 4 ern, respectively is

(v) In order to estimate the electric field due to a thin finite plane metal plate, the Gaussian surface considered is