This document discusses the application of Gauss's law to calculate electric field intensity in two situations: 1) Inside a hollow metallic sphere with charge distributed over its surface. Using Gauss's law, the electric field inside the sphere is found to be zero, showing how electric force can be shielded. 2) Near an infinite sheet with uniform surface charge density. A Gaussian cylinder is used to calculate the flux, yielding an electric field intensity of σ/(2ε0) perpendicular to the sheet. Gauss's law allows calculating electric fields for complex charge distributions.

1. PRESENTATION
Application of Gauss’s Law

2. SUBJECT : ELECTRICITY
AND MAGNETISM
PRESENTED BY : Muhammad Usama
BS Physics 3rd Semester
Roll.NO:10
PRESENTED TO: Sir Yamin

3. Introduction
Gauss’s law was formulated by German
Scientist Carl Friedrich Gauss in 1835.
Gauss’s Law is relating the distribution of
electric charge to result of electric field.

4. GAUSS’S LAW
Gauss’s Law states that:
“Electric flux through any Closed surface is equal to
1
𝜀°
time the total charge enclosed by that surface”
Φ =
1
𝜀°
Q

5. APPLICATION OF GAUSS’S
LAW
1. It is used to find electric field intensity at some
distance due to different charge distribution. over
single isolated charge we used E=k
𝑞
𝑟2 to find electric
field intensity but we can applied this formula for
multiple charge so we use Gauss’s law to find electric
field intensity more than one charge

6. STEP TO CALCULATE
ELECTRIC FIELD
INTENSITY
1. Draw a suitable Gaussian surface enclosing the
point where we want to calculate electric field
intensity .
Gaussian surface :
It is closed imaginary three dimensional surface

7. 2- calculate total electric flux using formula Φ= 𝑑𝐸.d𝐴 we
get equation (1)
3. Calculate electric flux using Gauss’s law Φ =
1
𝜀°
Q
We get equation (2)
4. Compare these two equation and we get electric field
intensity.

8. 1.ELECTRIC FIELD INTENSITY
INSIDE A HOLLOW SPHERE DUE
TO CHARGE DISTRIBUTED OVER
ITS SURFACE

9. EXPLANATION
Suppose we have
metallic sphere charge
distributed over the
surface. The charge is
reside over the
surface for metallic
object

10. The radius of sphere is R . Draw a Gaussian
surface in the form of sphere having radius R’ such
that R’<R.
The total electric flux through this closed surface
is calculated as
Φ= 𝑑𝐸. d𝐴
Φ= EA … . (1)

11. Applying Gauss’s law for close surface
Φ =
1
𝜀°
Q
As the interior of metallic sphere is hollow so
there is no electric charge Q=0
Φ = 0 … . . (2)

12. Compare equation (1) an(2) we get
𝐸𝐴 = 0
Α ≠ 0
𝐸 = 0
As Area is not zero its mean electric field intensity inside
the sphere is zero.

13. So E=0 its mean interior of metallic sphere is field
free region . Therefore we can say that electric force
can be shielded. That’s why all the sensitive
electronic devices are enclosed in metallic casing.
This application used in Tesla cage .

14. ELECTRIC FIELD
INTENSITY DUE TO
INFINITE SHEET OF
CHARGE

15. EXPLANATION
Consider a thin sheet of
infinite extent positive
charge is distributed over
its surface. We want to
calculate the electric field
intensity near the sheet at
point P .

16. Imagine a Gaussian surface in the form of cylinder of Area “A” take small segment
of area dA having charge dQ with surface charge density 𝜎
𝜎=
𝑑𝑄
𝑑𝐴
.dQ= 𝜎𝑑𝐴
∫dQ= ∫𝜎𝑑𝐴
Q= 𝜎𝐴
The Gaussian surface in the form of cylinder has three surfaces two flat ends P
and P’ and one curve end

17. FLUX THROUGH FIRST
FLAT END
Φ= 𝑑𝐸. d𝐴
Φ = 𝑑𝐸𝑑𝐴𝑐𝑜𝑠𝜃
𝜃 = 0
Φ= EA … . (1)

18. FLUX THROUGH FIRST
FLAT END
 Φ= 𝑑𝐸. d𝐴
 Φ = 𝑑𝐸𝑑𝐴𝑐𝑜𝑠𝜃
 𝜃 = 0
 Φ= EA … . (2)

19. ELECTRIC FLUX THROUGH
CURVE END
 Φ= 𝑑𝐸. d𝐴
 Φ = 𝑑𝐸𝑑𝐴𝑐𝑜𝑠𝜃
 𝜃 = 90
Φ=0 ……(3)


20. TOTAL ELECTRIC FLUX
Φ = EA + EA + 0
Φ = 2EA ……..(4)
From Gauss’s Law
Φ =
1
𝜀°
Q
Φ =
1
𝜀°
𝜎𝐴 ………(5)

21. Compare equation(4) and (5)
2EA=
1
𝜀°
𝜎𝐴
 E =
1
2𝜀°
𝜎
𝐸 =
𝜎
2𝜀°
𝑟

22. THANK YOU
The END