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 .

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.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>Jharkhand Board-2019>Q 3

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Two point charges of $+ 3 μC$ and $+ 4 μC$ repel each other with a force of 10 N. If each is given an additional charge of $- 6 μC$ , then the new force is.

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Important Questions on Electric Charges and Fields

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>Jharkhand Board-2019>Q 4

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

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>Jharkhand Board-2019>Q 5

An electric dipole with dipole moment $4 \times 10^{- 9} C m$ 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.