Write the statement of Gauss's law for electrostatics. Derive an expression for electric field due to an uniformly charged infinite non-conducting sheet at a point near to it. Draw suitable diagram.

Gauss Theorem: The net outward electric flux through a closed surface is equal to 1/ε0 times the net charge enclosed within the surface i.e., ∮ E → . d s → = 1 ε 0 ∑ q Let electric charge be uniformly distributed over the surface of a thin, non-conducting infinite sheet. Let the surface charge density (i.e., charge per unit surface area) be s. We have to calculate the electric field strength at any point distance r from the sheet of charge. To calculate the electric field strength near the sheet, we now consider a cylindrical Gaussian surface bounded by two plane faces A and B lying on the opposite sides and parallel to the charged sheet and the cylindrical surface perpendicular to the sheet (fig). By symmetry the electric field strength at every point on the flat surface is the same and its direction is normal outwards at the points on the two plane surfaces and parallel to the curved surface. Total electric flux. ∮ E → . d s → = ∮ E → . d s 1 → + ∮ E → . d s 2 → + ∮ E → . d s 3 → ∮ E → . d s → = ∫ E d s 1 cos 0 ° + ∫ E d s 2 cos 0 ° + ∫ E d s 3 cos 90 ° = E ∫ d s 1 + E ∫ d s 2 = E a + E a ∴ t o t a l e l e c t r i c f l u x = 2 E a As σ is charge per unit area of sheet and a is the intersecting area, the charge enclosed by Gaussian surface = σ a According to Gauss’s theorem, Total electric flux= 1 ε 0 σ a 2 E a = 1 ε 0 × σ a E = σ 2 ε 0 Thus, electric field strength due to an infinite flat sheet of charge is independent of the distance of the point and is directed normally away from the charge. If the surface charge density σ is negative the electric field is directed towards the surface charge.

Define: a Electric dipole moment.

An electric dipole is a separation of positive and negative charges. The electric dipole moment is the product of the magnitude of the charge and the distance between the centres of positive and negative charges. μ = Q × r Here, Q is the magnitude of charge and r is the distance between them.

Write definition of electric field intensity. Obtain an expression for electric force and electric pressure on the surface of a charged conductor. Draw necessary diagram.

Electric field intensity :- It is the strength of an electric field at any point. It is equal to the electric force per unit charge experienced by a test charge placed at that point. The unit of it i s volt/meter or Newton/ Coulomb. Every element of a charged conductor experienced a normal mechanical force. This is the result of repulsive force from similar change present on the rest of the surface of the conductor. Consider a small element ds on the surface of a charge conductor. if σ is the surface density and the charge carried by the element ds be dq. Consider a point P just outside the surface near the element ds. The electric intensity at a point P is given by E = σ ∈ 0 .................(1) The direction of the intensity is normally outwards. The intensity made up of two components E 1 → due to the small change dq present on the element ds. E 2 → due to the remaining charge present on the rest of the surface of the conductor. Hence, at P, E 1 → + E 2 → = σ ∈ 0 ..................(2) Now consider a point Q inside the conductor the electric field E 1 and E 2 at this point are oppositely directed. However, electric intensity inside a changed conductor is zero E 1 → - E 2 → = 0 E 1 → = E 2 → substituting E 1 in equation(2), we get, 2 E 2 = σ ∈ 0 E 2 = σ 2 ∈ 0 Here, E 2 is the electric field due to the charge on the rest of conductor. Hence, repulsive force experienced by elements ds carrying dq to is given by , f → = E 2 → . d q F = σ ∈ 0 . d q f = σ ∈ 0 . σ . d s F = σ 2 d s ∈ 0 The electric pressure on surface ds will be , pressure= force/area p r e s s u r e = σ 2 d s 2 ∈ 0 d s P = σ 2 2 ∈ 0 N / m 2