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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

Page1 / 12

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online