Jackson 2.3 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Transcription

1 Jackson.3 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell PROBLEM: A straight-line charge with constant linear charge λ is located perpendicular to the x-y plane in the first quadrant at (x, ). The intersecting planes at x, y and y, x 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, ) is Φ(x, y) (λ/4πε )ln( /r ), where r (x - x ) + (y - ) 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. (b) Determine the surface charge density σ on the plane y, x. Plot σ/λ versus x for (x, 1), (x 1, 1), (x 1, ). (c) Show that the total charge (per unit length in z) on the plane y, x is tan 1 x What is the total charge on the plane x? (d) Show that far from the origin [ρ >> ρ, where ρ x y and ρ x ] the leading term in the potential is asym 4 x xy 4 Interpret. SOLUTION: Using the method of images, let us put an image line charge λ' at (-x, ), an image line charge λ'' at (x, - ), and an image line charge λ''' at (-x, - ) and conceptually remove the conducting surface.

2 y y λ (x, ) (-x, ) λ' λ (x, ) x (-x, - ) λ''' λ'' (x, - ) x The solution to the potential for the four line charges is: x, y 1 ln 1 '' ln x x y 1 ' ln y 1 Apply the boundary condition x, y ln '''ln y x x y y ' ln y '' ln y ''' ln y R ln[ ' '' '''] y y y y R 1 ' '' ''' y y This can only be true for all y if ' and '' ''' Apply the boundary condition x, y ln ' ln x x R 1 R x x '' '''' xx ''ln This can only be true for all x if '' and ' ''' x x ''' ln xx

3 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 ln 1 ln y 1 x x y 1 ln ln x, y [ ln y lnxx y y xx y ln x x y ln y ] To explicitly verify that the potential disappears at the boundary, we check the potential at x x, y [ln x y ln x y lnx y ln x y ] x, y and check the potential at y x, y [ln x x lnxx ln x x ln ] x, y The tangential electric field along the x-axis boundary is just : x at y 4 x [ ln x x y ln y ln x x y ln y ] [ x x 4 y x x y x x y ] y

4 At y : [ 4 y x x y x x x x y xx ] at y : The tangential electric field along the y-axis boundary is just : y at x 4 y [ ln y lnx x y ln x x y ln y ] [ y 4 x x y y x x y y y y ] y At x : [ y 4 y y y y y at x y ] y (b) Determine the surface charge density σ on the plane y, x. Plot σ/λ versus x for (x, 1), (x 1, 1), (x 1, ). The surface charge density on an arbitrary surface creates an electric field discontinuity according to: [ E E 1 n 1 ]nn For a conductor, the electric field below the surface is zero, E 1, and the electric field is normal to the conductor's surface, and thus parallel to the conductor's normal, so that: [ E n 1 ]nn For this particular problem, the surface is the x-axis so that the normal is in the y direction [ ] y

5 We have already derived above and plug it directly in: [ [ y 4 x x y y x x y y x x y y xx y ]]y [ 4 4 y y 4 y ] 1 [ x x y 1 x x ] For (x, 1) 1 1 [ x 1 1 x 1 ] For (x 1, 1) 1 1 [ x x1 1 ]

6 For (x 1, ) 1 [ x x1 4] (c) Show that the total charge (per unit length in z) on the plane y, x is tan 1 x What is the total charge on the plane x? To get the total charge on the plane y we just integrate over the charge density on the plane: x dx [ [ x 1 y dx 1 x x y dx ] 1 x y dx x 1 x y dx ] [[ 1 1 tan x [ 1 1 tan x ] ] x ]x [ tan 1 x 1 tan 1 x 1 tan 1 x ]

7 Due to the total symmetry between the x and y axes, the total charge on the x plane is: tan 1 x (d) Show that far from the origin [ρ >> ρ, where ρ x y and ρ x ] the leading term in the potential is Interpret. asym 4 x xy 4 The potential was found above to be: x, y [ ln y lnxx y ln x x y ln y ] Put this potential in cylindrical coordinates (ρ, θ, z): x, y [ ln cos ln cos ln cos ln cos ] Divide everything by ρ so that we can get everything in terms of ρ /ρ and then we are able to make a statement about being far away from the origin: x, y [ ln ln 1 ln ln 1 x, y [ ln 1 ln 1 cos ln ln 1 cos ln ln 1 cos ln 1 cos ln 1 Expand each term in a Taylor series using ln 1xx x / x 3 /3... cos cos ] cos cos ]

8 x, y [[ [ ] [ cos 1/ cos ] 1/ [ [ [ cos ] 1/ [ cos ] 1/ [ Most of the first few terms cancel out when expanded: x, y 4 [ ] cos ] cos cos cos F 1 x 3, x 4,... F x 3, x 4,...F 3 x 3, x 4,...F 4 x 3, x 4,...] F 1 x 3, x 4,... F x 3, x 4,... ] cos ] cos F 3 x 3, x 4,... F 4 x 3, x 4,... ] Far away from the origin we have ρ >> ρ and therefore ρ /ρ<<1. This means that (ρ /ρ) 3 and (ρ /ρ) 4 etc. are negligible compared to (ρ /ρ) and they can all be dropped. x, y 4 cos cos x, y 4 coscos sin sin x, y 4 x x y 4 This is the quadrupole term, which makes sense because there are four line charges.