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

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

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

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

View Full Document
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 )
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

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

View Full Document
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
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 .

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

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

{[ snackBarMessage ]}

### 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
Ask a homework question - tutors are online