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

E M B I B E

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

EASY

12th Chhattisgarh Board

IMPORTANT

Earn 100

What is Electric Flux? Write its S.I. 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-2019>Q 5

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

HARD

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>chhattisgarh board-2019>Q 6

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

HARD

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Electric Charges and Fields>chhattisgarh board-2019>Q 7

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

EASY

12th Chhattisgarh Board

IMPORTANT

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

What is volume charge density? Write its SI unit.

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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