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Electric Field of a Ring
 Electric Field along axis of a Ring
 Electric Field along Axis of a disk
 Infinity Charged Plane
 Source and Reference

Electric Field of a Ring

image

Electric Field along axis of a Ring

image The electric field 𝐸 at ⟨0,0,π‘§βŸ© with radius π‘Ÿ due to charged particle at ⟨-π‘Ž,0,0⟩ is π‘Ÿ=βŸ¨π‘Ž,0,π‘§βŸ©β‡’|π‘Ÿ|=(π‘Ž2+𝑧2)1/2β‡’π‘Ÿ=π‘Ÿ|π‘Ÿ|=βŸ¨π‘Ž,0,π‘§βŸ©(π‘Ž2+𝑧2)1/2
𝐸 at ⟨0,0,π‘§βŸ©: 𝐸=14πœ‹πœ€0π‘ž|π‘Ÿ|2π‘Ÿ  =14πœ‹πœ€0π‘ž(π‘Ž2+𝑧2)βŸ¨π‘Ž,0,π‘§βŸ©(π‘Ž2+𝑧2)1/2=14πœ‹πœ€0π‘ž(π‘Ž2+𝑧2)3/2βŸ¨π‘Ž,0,π‘§βŸ©
The total electric field πΈπ‘‘π‘œπ‘‘ at ⟨0,0,π‘§βŸ© due to a ring of charged particles. By symmetry, the electric field πΈπ‘‘π‘œπ‘‘ πΈπ‘‘π‘œπ‘‘=β€‰β€‰βˆ‘aroundthe ringβˆ†πΈπ‘§π‘§
βˆ†πΈπ‘§=14πœ‹πœ€0π‘§βˆ†π‘ž(π‘Ž2+𝑧2)3/2
Let total net charge 𝑄 is uniformlly distributed on the ring with total length 𝐿 of radius π‘Ž. image The total electric field πΈπ‘‘π‘œπ‘‘ at ⟨0,0,π‘§βŸ© is 𝐿=2πœ‹π‘Ž; βˆ†π‘ž=𝑄2πœ‹π‘Žβˆ†πΏ=π‘„πΏβˆ†πΏ
βˆ†πΏ=π‘Žβˆ†πœƒβ‡’βˆ‘βˆ†πΏβ†’π‘Ž2πœ‹βˆ«0π‘‘πœƒ=2πœ‹π‘Ž=𝐿
β‡’βˆ†π‘ž=𝑄2πœ‹π‘Žπ‘Žβˆ†πœƒ=𝑄2πœ‹βˆ†πœƒ
β‡’βˆ‘βˆ†π‘žβ†’π‘„2πœ‹2πœ‹βˆ«0π‘‘πœƒ
β‡’πΈπ‘‘π‘œπ‘‘ 𝑧=14πœ‹πœ€0βˆ‘π‘§βˆ†π‘ž(π‘Ž2+𝑧2)3/2β†’14πœ‹πœ€0𝑄2πœ‹2πœ‹βˆ«0π‘§π‘‘πœƒ(π‘Ž2+𝑧2)3/2  =14πœ‹πœ€0𝑄2πœ‹π‘§(π‘Ž2+𝑧2)3/22πœ‹βˆ«0π‘‘πœƒ  =14πœ‹πœ€0𝑄𝑧(π‘Ž2+𝑧2)3/2

Electric Field along Axis of a disk

image The electric field 𝐸 at ⟨0,0,π‘§βŸ© with radius π‘Ÿ due to charged particle βˆ†π‘ž of βˆ†A=βˆ†π‘Žβˆ†πΏ at ⟨-π‘Ž,0,0⟩ is similar to that of a ring. image Let total net charge 𝑄 is uniformly distributed on the disk with total area 𝐴 of radius π‘Ž. Total 𝑄 in area A=πœ‹π‘Ž2
Elemental βˆ†π‘žπ‘– in area element βˆ†A𝑖=(π‘Žπ‘–βˆ†πœƒ)(βˆ†π‘Ž)
β‡’βˆ†π‘žπ‘–=π‘„πœ‹π‘Ž2βˆ†A𝑖=π‘„πœ‹π‘Ž2(π‘Žπ‘–βˆ†πœƒ)(βˆ†π‘Ž)
β‡’β€‰βˆ‘π‘–βˆ†π‘žπ‘–=β€‰βˆ‘π‘–π‘„πœ‹π‘Ž2βˆ†A𝑖=π‘„πœ‹π‘Ž2β€‰βˆ‘π‘–(π‘Žπ‘–βˆ†πœƒβˆ†π‘Ž)  =π‘„πœ‹π‘Ž2∬(π‘Žπ‘–π‘‘πœƒπ‘‘π‘Ž)=π‘„πœ‹π‘Ž2π‘Žβˆ«0π‘Žπ‘‘π‘Ž2πœ‹βˆ«0π‘‘πœƒ=π‘„πœ‹π‘Ž212π‘Ž22πœ‹=𝑄
β‡’πΈπ‘‘π‘œπ‘‘ 𝑧=14πœ‹πœ€0βˆ‘π‘§βˆ†π‘ž(π‘Ž2+𝑧2)3/2β†’14πœ‹πœ€0π‘„πœ‹π‘Ž2∬(π‘Žπ‘–π‘‘πœƒπ‘‘π‘Ž)𝑧(π‘Ž2+𝑧2)3/2  =14πœ‹πœ€0π‘„πœ‹π‘Ž2π‘Žβˆ«0π‘Žπ‘§(π‘Ž2+𝑧2)3/2π‘‘π‘Ž2πœ‹βˆ«0π‘‘πœƒ=14πœ‹πœ€0π‘„πœ‹π‘Ž21βˆ’π‘§π‘Ž2+𝑧22πœ‹  =12πœ€0π‘„πœ‹π‘Ž21βˆ’π‘§π‘Ž2+𝑧2 βˆ΄πΈπ‘‘π‘œπ‘‘ 𝑧=12πœ€0π‘„πœ‹π‘Ž21βˆ’π‘§π‘Ž2+𝑧2𝑧

Infinity Charged Plane

Assume π‘Ž go to infinity, e.g. π‘Žβ‰«π‘§ π‘Žβ†’βˆž; π‘„β†’βˆž β‡’π‘„πœ‹π‘Ž2=𝜎 charge per unit area
πΈπ‘‘π‘œπ‘‘ 𝑧=12πœ€0π‘„πœ‹π‘Ž21βˆ’π‘§π‘Ž2+𝑧2𝑧=12πœ€0𝜎1βˆ’0𝑧=𝜎2πœ€0𝑧
For two parallel opposite charged infinite plane image

Source and Reference

https://www.youtube.com/watch?v=WqSa620ln9M&list=PLZ6kagz8q0bvxaUKCe2RRvU_h7wtNNxxi&index=6


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ID: 191102202 Last Updated: 22/11/2019 Revision: 0


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