Gauss's Law and its applications

Gauss's law and its Application

1. presentation Application of Gauss’s Law

2. Subject : electricity and magnetism PRESENTED BY : Muhammad Usama BS Physics 3 rd 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 time the total charge enclosed by that surface”  

5. APPLICATION OF GAUSS’S law It is used to find electric field intensity at some distance due to different charge distribution. over single isolated charge we used E=k 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 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 = d we get equation (1) 3. Calculate electric flux using Gauss’s law 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  

11. Applying Gauss’s law for close surface As the interior of metallic sphere is hollow so there is no electric charge Q=0  

12. Compare equation (1) an(2) we get 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 = . dQ = ∫ dQ = ∫ Q= 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  

18. Flux through first flat end  

19. Electric flux through curve end =0 ……(3)  

20. Total electric flux = EA + EA + 0 = 2EA ……..(4) From Gauss’s Law ………(5)  

21. Compare equation(4) and (5) 2EA=  

22. Thank you The END