Chapter 2 - CHAPTER 2: Boundary-Value Problems in...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
region of interest region of 0 φ= 0 s s x x x y y σ x y q q q 0 0 image charge The method of images is not a general method. It works for some problems with a simple geometry. Consider a point charge located in front of an infinite and grounded plane conductor (see figure) q 2 0 . The region of interest is 0 and is governed by the Poisson equation: ( ) ( ) subject to the boundary condition ( 0) 0. In order to maintain a zero potential on the c q x x ε φ φδ ∇= == xx y onductor, surface charge will be induced (by ) on the conductor. We may simulate the effects of the surface charge with a hypothetical "image charge", , located symmetrically behind the conductor. T q q hen, 2.1 Method of Images CHAPTER 2: Boundary-Value Problems in Electrostatics: I
Background image of page 1

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

View Full DocumentRight Arrow Icon
region of interest region of 0 φ = 0 = s s x x x y y σ x y q q q 0 0 image charge 2 2 0 0 11 4 || | | ( ) and, by symmetry, ( ) satisfies the boundary condition ( 0) 0. Operate ( ) with ( ) [ ( ) ( )] q q x πε ε φδ δ −− ⎡⎤ =− ⎢⎥ ⎣⎦ == ⇒∇ xy xy x x x xx y x y 2 0 (1) In the region of interest ( 0), we have ( ) 0. Thus, ( ) obeys the original Poisson equation This shows that we must put the image ( ) ( ) ch q x ≥− = ∇= xy x y arge outside the region of interest Since ( ) satisfies both the Poisson equation and the boundary condition in the region of interest, it is a solution. By the uniqueness theorem, it is the x only solution. Note that the Poisson equation (1) and the solution ( ) are irrelevant outside the region of interest. x 2.1 Method of Images ( continued )
Background image of page 2
2.2 Point Charge in the Presence of a Grounded Conducting Sphere 00 44 Refer to the conducting sphere of radius shown in the figure. Assume a point charge is at ( ). To find for , we put an image charge at ( ). Then, () qq a q ry a r a qr y a πε φ − ′ => ′′ =< =+ xy x 2 2 Bounday condition requires 0 y a a y a y a y qa q a y −− = ⇒= nn n n x 2 First, set , or '= , ' so that = Note: ' ; hence, ' lies outside the region of interest. ' Next, set so that RHS 0. ' ' This gives ' . ya a y ay y y y q q a a = < = −= =− n n n B x q a y y n i q i
Background image of page 3

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

View Full DocumentRight Arrow Icon
2 2 2 00 0 44 This is equivalent to (2.1) Rewrite ( ) and (2.4) of Jackson. In the region of interest ( ), we have ( ) ( ). Thus, as in the case of the plane cond a y qa q q y ra π επ ε φ φδ =− ≥∇ = xy x xx y uctor, satisfies the Poisson equation and the b.c. It is hence the only solution. The -field lines are shown in the figure below. E 2.2 Point Charge in the Presence of a Grounded Conducting Sphere ( continued ) n B x q a y y n i q i
Background image of page 4
n B x q a y y n i q i 2 42 22 1 / 2 21 / 2 0 4 : The solution for ( ) can be expressed in terms of scalars as 1 ( ) (2 c o s ) c o s ) where is the angle between and [] q y y Surface charge density on the sphere a ax a xy x y yx πε φ γ =− +− x x x 2 00 2 43 3 / 2 23 / 2 8 .
Background image of page 5

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

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 28

Chapter 2 - CHAPTER 2: Boundary-Value Problems in...

This preview shows document pages 1 - 6. Sign up to view the full document.

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