Using Gauss's theorem, find this expression for electric field intensity at a distance r from a uniformly charged spherical shell of radius R in the following situations : r < R

Expression for electric field intensity due to a uniformly charged thin spherical shell for r < R i.e. when the point is considered inside the shell (internal point outside the shell) The spherical shell has a centre at O and radius R . Let the charge on the shell be q , it is uniformly distributed over the whole surface. Thus, the surface charge density σ = q A = q 4 π R 2 From centre O at distance r < R , P be a point where electric field intensity is to be determined. When point is considered inside the spherical shell and by taking O P = r , as a radius, a Gaussian surface is drawn then electric flux through Gaussian surface. = ϕ = ∮ E → ⋅ d A → = ∮ E d A c o s 0 ∘ = E ∮ d A = E . 4 π r 2 .....(i) Here no charge is enclosed by Gaussian surface because charges are the outer part of shell i.e., q i n = 0 According to Gauss's theorem: ϕ = ∮ E → ⋅ d A → = q ε 0 = 0 ϵ 0 = 0 ......(ii) From (i) and (ii), we get E ⋅ 4 π r 2 = 0 ⇒ E = 0

Using Gauss's theorem, find this expression for electric field intensity at a distance r from a uniformly charged spherical shell of radius R in the following situations : r = R

The spherical shell has a centre at O and radius R . Let the charge on the shell be q , it is uniformly distributed over the whole surface. Thus, the surface charge density σ = q A = q 4 π R 2 From centre O at distance r = R , P be a point where electric field intensity is to be determined. For this purpose distance O P = r , is taken as a radius and a spherical Gaussian surface is drawn. In this condition E → and d A → , are along the same direction. ∴ Electric flux = ϕ = E → ⋅ d A → = E d A cos 0 ° = E d A The electric flux through the whole Gaussian surface, ϕ = ∮ E → ⋅ d A → = ∮ E d A = E ⋅ ∮ d A = E × 4 π r 2 .....(i) Here charge q is enclosed by the Gaussian surface, so according to Gauss's theorem electric flux, ϕ = q ϵ 0 -----(ii) From equations (i) and (ii), we get E ⋅ 4 π r 2 = q ϵ 0 ⇒ E = 1 4 π ε 0 ⋅ q r 2 ----(iii) For r = R , Substituting the value in equation (iii) ∴ E 1 4 π ϵ 0 ⋅ q ( R ) 2 Thus, from above, E = q 4 π R 2 . 1 ϵ 0 = σ ϵ 0

Using Gauss's theorem, find this expression for electric field intensity at a distance r from a uniformly charged spherical shell of radius R in the following situations : r > R

Expression for electric field intensity due to a uniformly charged thin spherical shell for r > R i.e. when the point is considered outside the shell (external point outside the shell) The spherical shell has a centre at O and radius R . Let the charge on the shell be q , it is uniformly distributed over the whole surface. Thus, the surface charge density σ = q A = q 4 π R 2 From centre O at distance r > R , P be a point where electric field intensity is to be determined. For this purpose distance O P = r , is taken as a radius and a spherical Gaussian surface is drawn. In this condition E → and d A → , are along the same direction. ∴ Electric flux = ϕ = E → ⋅ d A → = E d A cos 0 ° = E d A The electric flux through the whole Gaussian surface, ϕ = ∮ E → ⋅ d A → = ∮ E d A = E ⋅ ∮ d A = E × 4 π r 2 .....(i) Here charge q is enclosed by the Gaussian surface, so according to Gauss's theorem electric flux, ϕ = q ϵ 0 -----(ii) From equations (i) and (ii), we get E ⋅ 4 π r 2 = q ϵ 0 ⇒ E = 1 4 π ε 0 ⋅ q r 2 = q 4 π R 2 × R 2 ε 0 r 2 = σ R 2 ε 0 r 2 ⇒ E ∝ 1 r 2

Using Gauss's law, derive the expression for electric field intensity due to a uniformly charged solid cylinder at a point that lies on the surface of the cylinder.

To calculate the electric field intensity due to a charged solid cylinder at a point r inside the cylinder, let us consider a solid cylinder of radius r and length equal to the length of the given cylinder to be the Gaussian surface. This surface is concentric with the given cylinder of radius R , length l and charge density ρ . From Gauss’s law, we have, ϕ = ∮ E → · d s → = q enclosed ε 0 ⇒ ϕ = q ε 0 = E · ∮ d s As the cylinder is infinitely long, an electric field will be along radial direction only, and hence flux through upper and lower caps of the Gaussian surface will be zero. ⇒ ρ π R 2 l ε 0 = E . 2 π r l ⇒ E = ρ R 2 2 ε 0 r For r = R , ∴ E = ρ R 2 ε 0

Using Gauss's law, derive the expression for electric field intensity due to a uniformly charged solid cylinder at a point that lies inside the cylinder.

To calculate the electric field intensity due to a charged solid cylinder at a point r inside the cylinder, let us consider a solid cylinder of radius r and length equal to the length of the given cylinder to be the Gaussian surface. This surface is concentric with the given cylinder of radius R , length l and charge density ρ . Thus, q = ρ × πr 2 l From Gauss’s law, we have, ϕ = ∮ E → · d s → = q enclosed ε 0 ⇒ ϕ = q ε 0 = E · ∮ d s As the cylinder is infinitely long, an electric field will be along radial direction only, and hence flux through upper and lower caps of the Gaussian surface will be zero. Thus, we will only include the area of curved surface of the cylinder. ⇒ ρ π r 2 l ε 0 = E . 2 π r l ⇒ E = ρ r 2 ε 0

Using Gauss's law, derive the expression for electric field intensity due to a uniformly charged solid cylinder at a point that lies outside the cylinder.

To calculate the electric field intensity due to a charged solid cylinder at a point r outside the cylinder, let us consider a solid cylinder to be the Gaussian surface which is concentric with the given cylinder of radius R . From Gauss’s law, we have, ϕ = ∮ E → · d s → = q enclosed ε 0 ⇒ ϕ = q ε 0 = E · ∮ d s As the cylinder is infinitely long, an electric field will be along radial direction only, and hence flux through upper and lower caps of the Gaussian surface will be zero. ⇒ ρ π R 2 l ε 0 = E . 2 π r l ⇒ E = ρ R 2 2 ε 0 r

What is Electric Flux? Write its S.I. unit.

Electric flux is the measure of the number of electric field lines passing through any surface. It can be defined as the measure of the distribution of the electric field or the rate of flow of the electric field through a given area. It is a scalar quantity. It is expressed using the Greek symbol ϕ . The SI base unit of electric flux is volt-meters ( Vm ), equal to newton meters squared per coulomb or ( Nm 2 C - 1 ).

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>chhattisgarh board-2020>Q 8

EASY

12th Chhattisgarh Board

IMPORTANT

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What is volume charge density? Write its SI unit.

Important Questions on Electric Charges and Fields

HARD

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>chhattisgarh board-2020>Q 9

Prove that the total electric flux associated with a closed surface is =

\frac{1}{ε_{0}}

times the total charge present inside the closed surface, where

ε_{0}

is the permittivity of free space. Write also the name of the theorem. What will be the value of electric flux if the charge lies outside the closed surface?

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