Homework3 - Homework 3 Answers, 95.657 Electromagnetic...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Homework 3 Answers, 95.657 Electromagnetic Theory I Dr. Christopher S. Baird, UMass Lowell Jackson 2.3 A straight-line charge with constant linear charge λ is located perpendicular to the x - y plane in the first quadrant at ( x 0 , y 0 ). The intersecting planes at x = 0, y ≥ 0 and y = 0, x ≥ 0 are conducting boundary surfaces held at zero potential. Consider the potential, fields, and surface charges in the first quadrant. (a) The well-known potential for an isolated line charge at ( x 0 , y 0 ) is Φ( x , y ) = (λ/4 πε 0 )ln( R 2 / r 2 ), where r 2 = ( x - x 0 ) 2 + ( y - y 0 ) 2 and R is a constant. Determine the expression for the potential of the line charge in the presence of the intersecting planes. Verify explicitly that the potential and the tangential electric field vanish at the boundary surfaces. SOLUTION: Using the method of images, let us put an image line charge λ ' at (- x 0 , y 0 ), an image line charge λ '' at ( x 0 , - y 0 ), and an image line charge λ ''' at (- x 0 , - y 0 ) and conceptually remove the conducting surface. The solution to the potential for the four line charges is:  x , y = 1 4  0 ln R 2 x x 0 2  y y 0 2 1 4  0 'ln R 2 x x 0 2  y y 0 2 1 4  0 ''ln R 2 x x 0 2  y y 0 2 1 4  0 '''ln R 2 x x 0 2  y y 0 2 Apply the boundary condition  x = 0, y = 0 0 = ln R 2 x 0 2  y y 0 2  'ln R 2 x 0 2  y y 0 2  ''ln R 2 x 0 2  y y 0 2  ''' ln R 2 x 0 2  y y 0 2 0 = ln [ R 2 x 0 2  y y 0 2 R 2 x 0 2  y y 0 2 ' R 2 x 0 2  y y 0 2 '' R 2 x 0 2  y y 0 2 ''' ] x y ( x 0 , y 0 ) λ x y ( x 0 , y 0 ) λ ( x 0 , - y 0 ) λ '' (- x 0 , y 0 ) λ ' (- x 0 , - y 0 ) λ '''
Background image of page 1

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

View Full DocumentRight Arrow Icon
1 = R 2 x 0 2  y y 0 2  ' R 2 x 0 2  y y 0 2 ''  ''' This can only be true for all y if  ' = 0 and ''  ''' = 0 Apply the boundary condition  x , y = 0 = 0 0 = ln R 2 x x 0 2 y 0 2  'ln R 2 x x 0 2 y 0 2  ''ln R 2 x x 0 2 y 0 2  ''' ln R 2 x x 0 2 y 0 2 1 = R 2 x x 0 2 y 0 2  '' R 2 x x 0 2 y 0 2 '  ''' This can only be true for all x if  '' = 0 and '  ''' = 0 Using the four equations in boxes, we now have four equations and three unknowns. We solve for each: ' =− , '' =− , ''' = The final solution is then:  x , y = 1 4  0 ln R 2 x x 0 2  y y 0 2 1 4  0 ln R 2 x x 0 2  y y 0 2 1 4  0 ln R 2 x x 0 2  y y 0 2 1 4  0 ln R 2 x x 0 2  y y 0
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/21/2012 for the course PHY 2030 taught by Professor Avery during the Spring '11 term at University of Florida.

Page1 / 18

Homework3 - Homework 3 Answers, 95.657 Electromagnetic...

This preview shows document pages 1 - 3. Sign up to view the full document.

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