An infinite line charge produces an electric field of 1 . 8 × 10 5 N C - 1 at a distance of 2 cm . Calculate the linear charge density.

Given, The electric field produced by the infinite linear charge E = 1 . 8 × 10 5 N C - 1 Distance at which electric field is calculated d = 2 cm = 2 × 10 - 2 m Electric field due to infinite linear charge at a distance d from it is E = λ 2 π ε ∘ d ⇒ λ = 2 π ε ∘ E d = 4 π ε ∘ E d 2 = 18 × 10 4 × 0 . 02 9 × 10 9 × 2 = 0 . 02 × 10 - 5 C m - 1 = 0 . 2 μ C m - 1 Hence, the linear charge density is 0 . 2 μ C m - 1 .

An infinite line charge produces an electric field of 1 . 8 × 10 5 N C - 1 at a distance of 2 cm . Calculate the linear charge density.

Given, The electric field produced by the infinite linear charge E = 1 . 8 × 10 5 N C - 1 Distance at which electric field is calculated d = 2 cm = 2 × 10 - 2 m Electric field due to infinite linear charge at a distance d from it is E = λ 2 π ε ∘ d ⇒ λ = 2 π ε ∘ E d = 4 π ε ∘ E d 2 = 18 × 10 4 × 0 . 02 9 × 10 9 × 2 = 0 . 02 × 10 - 5 C m - 1 = 0 . 2 μ C m - 1 Hence, the linear charge density is 0 . 2 μ C m - 1 .

An electric dipole with dipole moment 4 × 10 - 9 C m is inclined at 30 ° with the direction of a uniform electric field of magnitude 5×10 4 N C - 1 . Calculate the magnitude of torque acting on the dipole.

Electric dipole moment, p = 4 × 10 - 9 C m Angle made by dipole with the uniform electric field, θ = 30 ° Electric field, E = 5 × 10 4 N C - 1 Torque acting on the dipole is given by the relation, τ = p E s i n θ τ = 4 × 10 - 9 C m × 5 × 10 4 N C - 1 × sin 30 ° = 10 - 4 N m Hence, the magnitude of torque acting on the dipole is 10 - 4 N m .

State the principle of conservation and principle of quantization of electric charge.

Principle of conservation of electric charge: According to the law of conservation of electric charge, the net electric charge in an isolated system remain constant. Principle of quantization of electric charge: Existence of charges is discrete rather as continuous amount is known as quantization of electric charge. Quantization is the property of electric charge by virtue of which any charged body can have charge which is an integral multiple of elementary charge ‘ e ’ i.e q = n e .

State Gauss's theorem in electrostatics. Use this theorem to derive expressions for electric field due to uniformly charged thin spherical shell at-a point (i) outside and (ii) inside the shell.

Gauss's theorem in electrostatics: The total electric flux ϕ through any closed surface S in free space is equal to 1 ε ∘ times the total electric charge q enclosed by the surface. That is, ϕ = ∮ S E → . d s → = q ε ∘ , here ε ∘ is permittivity of free space. Electric field due to uniformly charged thin spherical shell at a point outside the shell: Here, O is the centre of the spherical shell of radius R , and charge enclosed by the spherical shell is q with uniformed charge density σ . Assuming the dotted circle of radius r to be Gaussian surface. Let a point P is outside the spherical shell. The distance between the centre of the spherical shell and the point P is r . Here, r > R . Now according to Gauss' theorem: ϕ = ∮ S E → . d s → = q ε ∘ ⇒ ϕ = ∮ S E d s cos θ = q ε ∘ Here, θ = 0 ° as E → and d s → are along the same direction. Thus, ϕ = ∮ S E d s = q ε ∘ ⇒ E ∮ S d s = q ε ∘ Since, ∮ S d s = 4 π r 2 Substituting this in above equation. E 4 π r 2 = q ε ∘ ⇒ E = 1 4 π ε ∘ q r 2 Hence, the expression for electric field at a point outside the uniformly charged thin spherical shell is E = 1 4 π ε ∘ q r 2 . Electric field due to uniformly charged thin spherical shell at a point inside the shell: When point P is inside the shell then r < R and inside the spherical shell charge q = 0 C . Then, E i n s i d e = 1 4 π ε ∘ q r 2 = 1 4 π ε ∘ 0 C r 2 = 0 N C - 1 .

Define electric dipole moment. Derive an expression for the electric field at a point on equatorial (broad-side-on) position of an electric dipole.

Electric dipole moment: The product of either change q of electric dipole and the distance between them l is referred to be Electric dipole moment. That is, p → = q × l . Electric field intensity on equatorial line of dipole: Considering an electric dipole with a pair of charge + q and - q separated by a small distance 2 a . Considering an equatorial line of dipole at a point P . At this point the net electric intensity will be due to charge + q and - q . Assuming E 1 → be the electric intensity at point P due to + q charge of the electric dipole. Then, E 1 → = 1 4 π ε ∘ q A P 2 along P C . Here, A P 2 = r 2 + a 2 by substituting this in above equation. E 1 → = 1 4 π ε ∘ q r 2 + a 2 ................................... 1 Here, ∠ P A B = ∠ P B A = θ Then the rectangular components of E 1 → are: Vertical component: E 1 → sin θ along P L Horizontal component: E 1 → sin θ along P R Assuming E 2 → be the electric intensity at point P due to - q charge of the electric dipole. Then, E 2 → = 1 4 π ε ∘ q P B 2 along P D . Here, P B 2 = r 2 + a 2 by substituting this in above equation. E 1 → = 1 4 π ε ∘ q r 2 + a 2 ................................... 2 Then the rectangular components of E 2 → are: Vertical component: E 2 → sin θ along P M Horizontal component: E 2 → cos θ along P R The net electric intensity at point P : E 1 → sin θ + E 1 → cos θ + E 2 → sin θ + E 2 → cos θ Here, the vertical component of E 1 → and E 2 → that is E 1 → sin θ and E 2 → sin θ are of same magnitude but in opposite direction so they will cancel out each other. So the net electric intensity at point P will be E n e t → = E 1 → cos θ + E 2 → cos θ = 2 E 1 → cos θ ⇒ E n e t → = 2 1 4 π ε ∘ q r 2 + a 2 cos θ Substituting, cos θ = a a 2 + r 2 in the above equation. E n e t → = 2 1 4 π ε ∘ q r 2 + a 2 a r 2 + a 2 = 1 4 π ε ∘ q 2 a r 2 + a 2 3 2 = 1 4 π ε ∘ p r 2 + a 2 3 2 Here, p is the electric dipole moment where p = q 2 a . E n e t → is along P R . If r > > a then E n e t → = 1 4 π ε ∘ p r 3 Hence, the expression for the electric field at a point on equatorial position of an electric dipole has been given.