Lesson_2.3_Printable_PPT

# Lesson_2.3_Printable_PPT - Electric Field Due to Extended...

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Unformatted text preview: Electric Field Due to Extended Charged Objects So far we discussed electric field due to point charges Gauss’s law will help us study electric fields due to extended charged objects. In order to study electric field due to extended charged objects, we will use the following terms: 1. Linear charge density ( λ ) This is the charge per unit length of a long charged object measured in Cm-1 If Q is the total charge on the object and L , its length: Q L λ = 2. Surface charge density ( σ ) This is the charge per unit area of a charged object measured in Cm-2 If Q is the total charge on the object and A , its area: Q A σ = . Volume charge density ( 3. Volume charge density ( ρ ) This is the charge per unit volume of a charged object measured in Cm-3 If Q is the total charge on the object and V , its volume: Q V ρ = Consider a long charged object of length L and linear charge density λ with one end at the origin O O L P x dx x E on the axis of a finite line charge. We take a small segment of length dx at a distance x from the origin so that for all practical purposes, it can be considered to be a point charge and Coulomb’s law can be applied. We want to find the electric field at P a distance x from the origin. The charge on the small segment of length dx is λ dx The field dE at P due the charge on dx is O L P x dx x kq E k dx λ dE The field E at P due to the entire object can be obtained by integrating dE from x = 0 to x = L 2 dE r = 2 ( ) x x =- 2 ( ) L k dx E x x λ =- ∫ 2 ( ) L k dx E x x λ =- ∫ 2 1 ( ) L k dx x x λ =- ∫ 1 ( ) L k x x λ = - 1 1 ( ) ( 0) k x L x λ =- -- ( ) ) x x L k x L λ -- = ( ) x x L- ( ) L k x x L λ = - ( ) k x x L L λ =- ( ) kQ x x L =- When P is very far, x – L ≈ x . 2 kQ E x = At points very far from the object, the object behaves like a point charge Example 1: A uniform line charge of charge density λ = 3.5 nC.m-1 extends from x = 0 to x = 5.0 m. (a) What is the total charge? Find the electric field at (b) x = 6.0 m, (c) x = 9.0 m and (d) x = 250 m λ = 3.5 nCm-1 L = 5 m (a) Q = L λ = 5 m × 3.5 nCm-1 = 17.5 nC (b) To find field at x = 6.0 m, we use x = 6 m ( ) kQ E x x L =- 9 9 9 10 17.5 10 6(6 5)- × × × =- = 26.25 NC-1 Example 1: A uniform line charge of charge density λ = 3.5 nC.m-1 extends from x = 0 to x = 5.0 m. (a) What is the total charge? Find the electric field at (b) x = 6.0 m, (c) x = 9.0 m and (d) x = 250 m λ = 3.5 nCm-1 L = 5 m (c) To find field at x = 6.0 m, we use x = 9 m kQ 9 9 9 10 17.5 10- × × × 4.4 NC-1 ( ) E x x L =- 9(9 5) =- = 4.4 NC (d) x = 250 is very large compared to L = 5. We use the approximate formula for large distances....
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Lesson_2.3_Printable_PPT - Electric Field Due to Extended...

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