Electric field due to ring of charge Derivation

This is derivation of physics about electric field due to a charged ring.This is complete expression.

1. Electric field due to ring & disc of charge

2. Electric Field due to Ring of Charge Consider a positively charged ring having radius R on which positive charge q is distributed uniformly. This is called linear charge distribution. Take a small length element ds of ring having charge dq . The linear charge density λ is defined as λ Now consider two length elements ds at opposite ends of a diameter of ring. We have to calculate electric field at P which is now perpendicular axis at distance z from the ring.   2

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4. Electric Field due to Ring of Charge From figure: The magnitude of electric field at P due to charge element L is Similarly, the magnitude of electric field at P due to charge element M is   4

5. Electric Field due to Ring of Charge Comparing both equations, we get Now calculate the electric field at P due to both length elements, resolve the electric field and into components. Rectangular components of are , Rectangular components of are ,   5

6. Electric Field due to Ring of Charge Resultant x-component of electric field is   6 As  

7. Electric Field due to Ring of Charge Resultant y-component of electric field is   7 As  

8. Electric Field due to Ring of Charge The magnitude of electric field at P due to whole ring of charge is   8 As   From λ  

9. Electric Field due to Ring of Charge As, = Length of circumference of half ring because length elements are taken at both sides of diameter. As λ So q = λ = λ   9

10. Electric Field due to Ring of Charge   10

11. Electric Field due to Ring of Charge This is the electric field at point P due to ring of charge. When point P lies at a large distance z. the term can be neglected as compared to   It means that ring behaves as a point charge which is concentrated at center of ring when z>>R.   11

12. Electric Field due to Disc of Charge Consider a positively charged circular disc of radius R having surface charge density D𝑖𝑣𝑖𝑑𝑒 𝑡ℎ𝑒 𝑑𝑖𝑠𝑐 𝑖𝑛𝑡𝑜 𝑠𝑚𝑎𝑙𝑙 𝑟𝑖𝑛𝑔𝑠. 𝑛𝑜𝑤 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑢𝑐ℎ 𝑎 𝑠𝑚𝑎𝑙𝑙 𝑟𝑖𝑛𝑔 ha𝑣𝑖𝑛𝑔 𝑟𝑎𝑑𝑖𝑢𝑠 ω and d ω T ake a pair of small area element da at opposite ends of diameter denoted by L and M. the area of the element having width d ω and length S= ω d α is given as dA = ω d α d ω   12

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14. Electric Field due to Disc of Charge The charge per unit area is called surface charge density = d q = dA dq = ( ω d α d ω )   14

15. Electric Field due to Disc of Charge The magnitude of electric field at P due to charge element L is Similarly, the magnitude of electric field at P due to charge element M is   15

16. Electric Field due to Disc of Charge Comparing both equations, we get Now calculate the electric field at P due t both length elements, resolve the electric field and into components. Rectangular components of are , Rectangular components of are ,   16

17. Electric Field due to Disc of Charge Resultant x-component of electric field is   17 As  

18. Electric Field due to Disc of Charge Resultant y-component of electric field is   18 As  

19. Electric Field due to Disc of Charge The magnitude of electric field dE is + dE = dq = ( ω d α d ω ) and   19

20. Electric Field due to Disc of Charge The magnitude of electric field at P due to whole ring of charge is   20

21. Electric Field due to Disc of Charge   21

22. Electric Field due to Disc of Charge 22

23. Electric Field due to Disc of Charge 23