Lesson_2.3

# Lesson_2.3 - Lesson 2.3 Electric Field due to continuous...

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Unformatted text preview: Lesson 2.3 Electric Field due to continuous charge distributions 1. Field Calculations using Coulomb’s law In this section we will use the following terms: λ = linear charge density which is charge per unit length. Q L λ = σ = surface charge density and is the charge per unit area. Q A σ = ρ = volume charge density and is charge per unit volume. Q V ρ = (i) E on the axis of a finite line charge. + + + + + . P dx dq = λ dx L Fig. 1 x o x Coulomb’s law can be applied only to point charges and not for extended charges. Therefore, to find the electric field E at P due to the extended charge of length L, 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. If λ is the charge per unit length of the line charge, the charge dq of the small segment of length dx is given by dq = λ dx If x o is the distance of P from the origin, the field at P due to dq is 2 2 ( ) ( ) kdq k dx dE x x x x λ = =-- Since the position of dx can change from x = 0 to x = L, the total field E at P due to the line charge is 2 1 1 ( ) 1 1 ( ) L L E k dx k x x x x L k k x L x x x L λ λ λ λ = = -- =- = -- ∫ But λ L = Q, the total charge on the line charge. Therefore, ( ) x kQ E x x L =- …1 when x o is much larger than L, this becomes: 2 x kQ E x = …2 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 Solution: (a) Total charge Q = L λ = 5.0 × 3.5 × 10-9 = 17.5 nC (b) 9 1 9 9.0 10 17.5 26.25 10 ( ) 6(6 5) x kQ E x L N x C-- × × × = = =-- (c) 9 1 9 9.0 10 17.5 10 ( ) 9(9 5) 4.38 x kQ E x x NC L-- × × × = = =-- (d) Since x o = 250 m is much larger than L = 5.0 m, we can use the approximate formula for E x here. 9 9 2 2 3 1 9.0 10 17.5 1 2. 1 2 5 5 x kQ E x N C--- × × × = = = × ⋅ (ii) Field on the axis of a ring charge. Fig. 2 below shows a ring of radius r carrying a charge Q and its center at the origin. The point P is on the x axis at a distance x from the origin. In order to find the field at P due to the charge on the ring, we take a small element of the ring which has a charge dq. This charge dq is at a distance r from P such that r 2 = x 2 + a 2 where a is the radius of the ring. . P a x r θ dE dE x dE p dE p Fig. 2 dq θ The field dE at P due to the charge dq is 2 kdQ dE r = This field dE has a component dE cos θ (dE x ) along the x axis and a component dE sin θ (dE p ) at right angles to the x axis. By symmetry, the components at right angles to the x axis will all cancel and only dE x , the component along the x axis will remain....
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Lesson_2.3 - Lesson 2.3 Electric Field due to continuous...

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