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Application of Gauss's Law

1. The document discusses Gauss's law and its application to calculate electric fields due to various charge distributions. 2. It first calculates the electric field due to an infinite line charge, showing the field is proportional to the linear charge density divided by distance from the line. 3. It then calculates the electric field due to an infinite plane sheet, showing the field is proportional to the surface charge density divided by the permittivity of free space. 4. Finally, it calculates the electric field due to a uniformly charged thin spherical shell, deriving expressions for the field inside, outside, and on the surface of the shell based on Gauss's law.

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Sanjay
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6K views6 pages

Application of Gauss's Law

1. The document discusses Gauss's law and its application to calculate electric fields due to various charge distributions. 2. It first calculates the electric field due to an infinite line charge, showing the field is proportional to the linear charge density divided by distance from the line. 3. It then calculates the electric field due to an infinite plane sheet, showing the field is proportional to the surface charge density divided by the permittivity of free space. 4. Finally, it calculates the electric field due to a uniformly charged thin spherical shell, deriving expressions for the field inside, outside, and on the surface of the shell based on Gauss's law.

Uploaded by

Sanjay
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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1 Physics Chapter 1

APPLICATION OF GAUSS’S LAW

FIELD DUE TO AN INFINITY LONG CHARGED WIRE


Consider a thin infinity long straight wire having a uniform linear charge density λ. We
choose a cylindrical Gaussian surface of radius r , length l and width its axis along the line
charge. Let S1 be the curved surface and S2 and S3 be the flat ends. Only the curved surface
contributes towards the total flux.

∅ = ∮ 𝐸⃗ ∙ ⃗⃗⃗⃗
𝑑𝑠 + ∮ 𝐸⃗ ∙ ⃗⃗⃗⃗
𝑑𝑠 + ∮ 𝐸⃗ ∙ ⃗⃗⃗⃗
𝑑𝑠
𝑠1 𝑠2 𝑠3

∅ = ∮ 𝐸 𝑑𝑆1 cos 0° + ∮ 𝐸 𝑑𝑆2 cos 90° + ∮ 𝐸 𝑑𝑆3 cos 90°


𝑠1 𝑠2 𝑠3

∅ = 𝐸 ∮ 𝑑𝑠1
𝑠1

∅ = 𝐸. 2𝜋𝑟𝑙 ---------①

Charge enclosed by the Gaussian surface , q= λl

Using Gauss’s theorem


𝑞 λl
∅=𝜀 =𝜀 ------------②
𝑜 𝑜

A notes by collegestudynotes.com
2 Physics Chapter 1

Equating ②&① we get


λl
𝐸. 2𝜋𝑟𝑙 = 𝜀
𝑜

λ
𝐸=
2𝜋𝜀𝑜 𝑟

ELECTRIC FIELD DUE TO UNIFORMLY CHARGED INFINITE PLANE SHEET


Consider a thin, infinite plane sheet of charge with uniform surface charge density σ. Let P
be a point at distance r from it. We choose cylindrical Gaussian surface of cross sectional
area A and length 2r with its axis perpendicular to the sheet.

As the field lines are parallel to the curved surface of the cylinder, the flux through the
curved surface is zero. The flux through the plane end faces of the cylinder is

Φ =2EA -----1

Change enclosed by the Gaussian surface , q = σA

According to gauss’s theorem


𝑞 𝜎𝐴
∅=𝜀 = 𝜀𝑜
------2
𝑜

Equating eq. 1 and 2 , we get

𝜎𝐴
2𝐸𝐴 =
𝜀𝑜

A notes by collegestudynotes.com
3 Physics Chapter 1

𝜎
𝐸=
2𝜀𝑜

NOTE: (i) If the sheet is positively charged the field is directed away from
it.
(ii) If the sheet is negatively charged the field is directed towards it.

FIELD DUE TO A UNIFORMLY CHARGED THIN SPHERICAL SHELL


Consider a thin spherical shell of radius R and uniform surface charge density σ. Let P be a
point at a distance r from O we choose a concentric sphere of radius r as the Gaussian
surface.

CASE 1 When P is outside the shell


The total charge q inside the Gaussian surface is the charge on the shell of radius R and
area 4πR2

q = 4πR2 σ

A notes by collegestudynotes.com
4 Physics Chapter 1

flux through the Gaussian surface

∅ = 𝑬. 𝟒𝝅𝒓𝟐 ------1
By gauss’s law

𝑞 4𝜋𝑟 2
∅= = ------2
𝜀𝑜 𝜀𝑜

On equating eq. 1 & 2 we get

4𝜋𝑅 2 𝜎
𝐸. 4𝜋𝑟 2 =
𝜀𝑜
𝑅2𝜎 1 𝑞
𝐸= OR 𝐸=
𝑟 2𝜀 𝑜 4𝜋𝜀𝑜 𝑟 2

NOTE: The field is same as that produced by a charge q placed at the centre O. hence
for all points outside the shell the field due to uniformly charged shell is as if the entire
charge of the shell is concentrated at its centre.

CASE 2 When point P lies on the spherical shell.


The Gaussian surface just encloses the charged spherical shell.

Applying gauss’s theorem

𝑞
𝐸. 4𝜋𝑅 2 =
𝜀𝑜
1 𝑞
𝐸=
4𝜋𝜀𝑜 𝑅2
𝜎
𝐸=
𝜀𝑜

CASE 3 When point p lies inside the spherical shell


In this case , the charge enclosed by the Gaussian surface is zero.

Flux through Gaussian surface

A notes by collegestudynotes.com
5 Physics Chapter 1

∅ = 𝐸. 4𝜋𝑟 2 -------1

Applying Gauss’s theorem


𝑞
∅= =0 -----2
𝜀𝑜

Equating eqn. 1 & 2 we get,

E=0

NOTE: Electric field due to uniformly charged spherical shell is zero at all points inside the
shell.

Variation of electric field ‘E’ with distance r from the centre of the shell

A notes by collegestudynotes.com
6 Physics Chapter 1

A notes by collegestudynotes.com

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