Using Gauss’s theorem in electrostatics, derive an expression for electric field at a point due to an infinitely long line of charge having a uniform charge density.

According to Gauss's theorem, the net outward electric flux through a closed surface is equal to 1 ε ∘ times the net charge enclosed within the surface. That is, ∮ S E → . d s → = 1 ε ∘ ∑ q Now, for electric field at a point due to an infinitely long line of charge having a uniform charge density λ then for the l length the net charge will be ∑ q = λ l . Then, the electric field will be: ∮ S E → . d s → = 1 ε ∘ λ l ⇒ E 2 π r l = 1 ε ∘ λ l ⇒ E = λ 2 π r ε ∘ Hence, the electric field at a point due to an infinitely long line of charge having a uniform charge density is λ 2 π r ε ∘ .

Obtain an expression for the torque on an electric dipole placed in a uniform electric field.

Consider an electric dipole of length 2 d , having dipole moment p → placed in a uniform electric field E → . The dipole is placed at an angle θ . Since the dipole is in uniform electric field thus, both the charges of the dipole will experience force of F = q E . As both the charges of the dipole experience equal and opposite force the dipole will start rotating. Thus, the torque experienced by the electric dipole: τ = q E 2 d sin θ τ = 2 d q E sin θ = p E sin θ = p → × E → Hence, the torque on an electric dipole placed in a uniform electric field is p → × E → .

Use Gauss’s theorem to derive an expression for electric field due to a uniformly charged spherical shell at a point outside the shell.

Consider a charged spherical shell of radius R on which Q charge is distributed uniformly over the surface. Assuming the surface of the spherical shell be a Gaussian surface. Then, the flux enclosed by the charged spherical shell is: ϕ = ∮ E . d s = Q ε ∘ ⇒ E × 4 π R 2 = Q ε ∘ ⇒ E = 1 4 π R 2 Q ε ∘ So, for a point outside the spherical shell the distance between the centre of the spherical shell and the outside point is r where, r > R . Then, the field outside the shell will be: E o u t s i d e = 1 4 π r 2 Q ε ∘ .

S 1 and S 2 are two parallel concentric spheres enclosing charges Q and 2 Q respectively as shown in the figure below. What is the ratio of electric flux through S 1 and S 2 ?

The flux through S 1 : ϕ 1 = Q ε ∘ The flux through S 2 : ϕ 2 = Q + 2 Q ε ∘ = 3 Q ε ∘ So, the ratio of the flux: ϕ 1 ϕ 2 = Q ε ∘ 3 Q ε ∘ = 1 3 = 1 : 3 Hence, the ratio of the flux through S 1 and S 2 is 1 : 3 .