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 =
𝑑𝑞
𝑑𝑠
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

3. 3

4. Electric Field due to Ring of Charge
From figure:
𝑟2
= 𝑧2
+ 𝑅2
The magnitude of electric field at P due to charge
element L is
𝑑𝐸𝐿 =
𝑘 𝑑𝑞
𝑟2
Similarly, the magnitude of electric field at P due to
charge element M is
𝑑𝐸 𝑀 =
𝑘 𝑑𝑞
𝑟2
4

5. Electric Field due to Ring of Charge
Comparing both equations, we get
𝑑𝐸𝐿 = 𝑑𝐸 𝑀
Now calculate the electric field at P due to both length
elements, resolve the electric field 𝑑𝐸𝐿 and 𝑑𝐸 𝑀 into
components.
Rectangular components of 𝑑𝐸𝐿 are
𝑑𝐸𝐿𝑥 = 𝑑𝐸𝐿 cos θ , 𝑑𝐸𝐿𝑦 = 𝑑𝐸𝐿 sin θ
Rectangular components of 𝑑𝐸 𝑀 are
𝑑𝐸 𝑀𝑥 = 𝑑𝐸 𝑀 cos θ , 𝑑𝐸 𝑀𝑦 = 𝑑𝐸 𝑀 sin θ
5

6. Electric Field due to Ring of Charge
Resultant x-component of electric field is
𝑑𝐸 𝑥 = (+𝑑𝐸𝐿𝑥) + (−𝑑𝐸 𝑀𝑥)
𝑑𝐸 𝑥 = 𝑑𝐸𝐿 cos θ − 𝑑𝐸 𝑀 cos θ
𝑑𝐸 𝑥 = 𝑑𝐸𝐿 cos θ − 𝑑𝐸𝐿 cos θ
𝑑𝐸 𝑥 = 0
6
As 𝑑𝐸𝐿 = 𝑑𝐸 𝑀

7. Electric Field due to Ring of Charge
Resultant y-component of electric
field is
𝑑𝐸 𝑦 = (+𝑑𝐸𝐿𝑦) + (+𝑑𝐸 𝑀𝑦)
𝑑𝐸 𝑦 = 𝑑𝐸𝐿 𝑠𝑖𝑛 𝜃 + 𝑑𝐸 𝑀 𝑠𝑖𝑛 𝜃
𝑑𝐸 𝑦 = 𝑑𝐸𝐿 𝑠𝑖𝑛 𝜃 + 𝑑𝐸𝐿 𝑠𝑖𝑛 𝜃
𝑑𝐸 𝑦 = 2 𝑑𝐸𝐿 𝑠𝑖𝑛 𝜃
7
As 𝑑𝐸𝐿 = 𝑑𝐸 𝑀

8. Electric Field due to Ring of Charge
The magnitude of electric field at P due to whole ring of charge is
𝐸 = 2 𝑑𝐸𝐿 sin 𝜃
𝐸 = 2
𝑘 𝑑𝑞
𝑟2
.
𝑧
𝑟
𝐸 = 2
𝑧 𝑘 𝑑𝑞
𝑟3
𝐸 = 2
𝑧 𝑘 λ
𝑟3
𝑑𝑠
𝐸 = 2
𝑧 𝑘 λ
(𝑧2 + 𝑅2)
3
2
𝑑𝑠
8
As sin 𝜃 =
𝑧
𝑟
From λ =
𝑑𝑞
𝑑𝑠

9. Electric Field due to Ring of Charge
As, 𝑑𝑠 = 𝜋 𝑅
Length of circumference of half ring because length elements are taken
at both sides of diameter.
𝐸 =
2 𝑧 𝑘 λ
𝑧2 + 𝑅2 3
2
(𝜋 𝑅)
As λ =
𝑑𝑞
𝑑𝑠
So q = λ 𝑑𝑠 = λ(2 𝜋 𝑅)
9

10. Electric Field due to Ring of Charge
𝐸 =
2 𝑧 𝑘 λ
𝑧2 + 𝑅2 3
2
(𝜋 𝑅)
𝐸 =
𝑧 𝑘 λ
𝑧2 + 𝑅2 3
2
(2 𝜋 𝑅)
𝐸 =
𝑧 𝑘 q
𝑧2 + 𝑅2 3
2
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 𝑅2
can be
neglected as compared to 𝑧2
𝐸 =
𝑧 𝑘 q
𝑧2 + 𝑅2 3
2
𝐸 =
𝑧 𝑘 q
𝑧2 + 0
3
2
𝐸 =
𝑧 𝑘 q
𝑧2 3
2
𝐸 =
𝑧 𝑘 q
𝑧3
𝐸 =
𝑘 q
𝑧2
It means that ring behaves as a point
charge which is concentrated at center of
ring when z>>R.
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12. Electric Field due to Disc of Charge
Consider a positively charged circular disc of radius R
having surface charge density 𝜎.
D𝑖𝑣𝑖𝑑𝑒 𝑡ℎ𝑒 𝑑𝑖𝑠𝑐 𝑖𝑛𝑡𝑜 𝑠𝑚𝑎𝑙𝑙 𝑟𝑖𝑛𝑔𝑠. 𝑛𝑜𝑤 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑢𝑐
ℎ 𝑎 𝑠𝑚𝑎𝑙𝑙 𝑟𝑖𝑛𝑔 ha𝑣𝑖𝑛𝑔 𝑟𝑎𝑑𝑖𝑢𝑠 ω and dω
Take a pair of small area element da at opposite ends
of diameter denoted by L and M. the area of the
element having width dω and length S=ωdα is given as
dA= ω dα dω
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13. 13

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

15. Electric Field due to Disc of Charge
The magnitude of electric field at P due to charge
element L is
𝑑𝐸𝐿 =
𝑘 𝑑𝑞
𝑟2
Similarly, the magnitude of electric field at P due to
charge element M is
𝑑𝐸 𝑀 =
𝑘 𝑑𝑞
𝑟2
15

16. Electric Field due to Disc of Charge
Comparing both equations, we get
𝑑𝐸𝐿 = 𝑑𝐸 𝑀
Now calculate the electric field at P due t both length
elements, resolve the electric field 𝑑𝐸𝐿 and 𝑑𝐸 𝑀 into
components.
Rectangular components of 𝑑𝐸𝐿 are
𝑑𝐸𝐿𝑥 = 𝑑𝐸𝐿 cos θ , 𝑑𝐸𝐿𝑦 = 𝑑𝐸𝐿 sin θ
Rectangular components of 𝑑𝐸 𝑀 are
𝑑𝐸 𝑀𝑥 = 𝑑𝐸 𝑀 cos θ , 𝑑𝐸 𝑀𝑦 = 𝑑𝐸 𝑀 sin θ
16

17. Electric Field due to Disc of Charge
Resultant x-component of electric field is
𝑑𝐸 𝑥 = (+𝑑𝐸𝐿𝑥) + (−𝑑𝐸 𝑀𝑥)
𝑑𝐸 𝑥 = 𝑑𝐸𝐿 cos θ − 𝑑𝐸 𝑀 cos θ
𝑑𝐸 𝑥 = 𝑑𝐸𝐿 cos θ − 𝑑𝐸𝐿 cos θ
𝑑𝐸 𝑥 = 0
17
As 𝑑𝐸𝐿 = 𝑑𝐸 𝑀

18. Electric Field due to Disc of Charge
Resultant y-component of electric field is
𝑑𝐸 𝑦 = (+𝑑𝐸𝐿𝑦) + (+𝑑𝐸 𝑀𝑦)
𝑑𝐸 𝑦 = 𝑑𝐸𝐿 𝑠𝑖𝑛 𝜃 + 𝑑𝐸 𝑀 𝑠𝑖𝑛 𝜃
𝑑𝐸 𝑦 = 𝑑𝐸𝐿 𝑠𝑖𝑛 𝜃 + 𝑑𝐸𝐿 𝑠𝑖𝑛 𝜃
𝑑𝐸 𝑦 = 2 𝑑𝐸𝐿 𝑠𝑖𝑛 𝜃
18
As 𝑑𝐸𝐿 = 𝑑𝐸 𝑀

19. Electric Field due to Disc of Charge
The magnitude of electric field dE is
𝑑𝐸2
= 𝑑𝐸 𝑥
2
+ 𝑑𝐸 𝑦
2
𝑑𝐸 = 2 𝑑𝐸𝐿 𝑠𝑖𝑛 𝜃
dE = 2
𝑧 𝑘 𝑑𝑞
𝑟3
dq = 𝜎 (ω dα dω) and 𝑟2
= (𝑧2
+ ω2
)
19

20. Electric Field due to Disc of Charge
The magnitude of electric field at P due to whole ring of charge is
𝐸 = 𝑑𝐸
𝐸 = 2
𝑧 𝑘 𝑑𝑞
𝑟3
𝐸 = 2
𝑧 𝑘 𝜎 (ω dα dω)
(𝑧2 + ω2)
3/2
𝐸 = 𝑧 𝑘 𝜎 2
(ω dα dω)
(𝑧2 + ω2)
3/2
𝐸 = 𝑧 𝑘 𝜎
0
𝑅
2
(ω dω)
(𝑧2 + ω2)
3/2
0
𝜋
dα
20

21. Electric Field due to Disc of Charge
𝐸 = 𝑧 𝑘 𝜎
0
𝑅
2
(ω dω)
(𝑧2 + ω2)
3/2
0
𝜋
dα
𝐸 = 𝑧 𝑘 𝜎
0
𝑅
(𝑧2 + ω2)
−3/2
(2 ω dω) ( 𝜋−0)
𝐸 = 𝑧 𝑘 𝜎 𝜋
0
𝑅
(𝑧2
+ ω2
)
−3/2
(2 ω dω)
21

22. Electric Field due to Disc of Charge
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23. Electric Field due to Disc of Charge
23