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Jackson 4.5 Homework Problem Solution
Dr. Christopher S. Baird University of Massachusetts Lowell PROBLEM: A localized charge density ρ( x , y , z ) is placed in an external electrostatic field described by a potential Φ(^0 )( x , y , z ). The external potential varies slowly in space over a region where the charge density is different from zero. (a) From first principles calculate the total force acting on the charge distribution as an expansion in multipole moments times derivatives of the electric field, up to and including the quadrupole moments. Show that the force is F = q E ( 0 ) +(∇ [ p ⋅ E ( 0 ) ]) 0 + (
[
∑ j , k Q (^) jk ∂ E (^) j ( 0 ) ∂ xk ( x ) ]) 0
Compare this to the expansion (4.24) of the energy W. Note that (4.24) is a number – it is not a function of x that can be differentiated! What is its connection to F? (b) Repeat the calculation of part a for the total torque. For simplicity, evaluate only one Cartesian component of the torque, say N 1. Show that this component is: N (^) 1 =[ p × E ( 0 ) ( 0 )] 1 +
3 [
∂ x 3 ( ∑ j Q 2 j E (^) j ( 0 ) )
∂ x 2 ( ∑ j Q 3 j E (^) j ( 0 ) )] 0
SOLUTION:
(a) The force in general is: F =∫ρ( x ) E ( 0 ) ( x ) d 3 x Let us keep track of the components. F =∑ i
x i ∫ρ( x ) Ei ( 0 ) ( x ) d 3 x Expand the external electric field in a Taylor series and only keep the focus on the first few terms because it varies slowly over space: Ei ( 0 ) ( x )= [ Ei ( 0 ) ( x ')+∑ j x (^) j
∂ x (^) j ' Ei ( 0 ) ( x ')+
∑ j , k x (^) j xk
∂ x (^) j '
∂ xk ' Ei ( 0 ) ( x ')+... ] x '= 0 Note we have labeled the gradients as being done with respect to primed variables to tell them apart from the integration variables. Substitute the expansion into the force equation:
F =
[
E
( 0 ) ( x ') q +∑ i ∑ j
x i ∫ρ( x ) x (^) j d 3 x
∂ x (^) j ' Ei ( 0 )
∑ i ∑ j , k
x i ∫ρ( x ) x (^) j xk d 3 x
∂ x (^) j '
∂ xk ' Ei +...
] x '= 0
Note that ∇× E =^0 means that ∂ Ei ∂ x (^) j
∂ E (^) j ∂ xi
. We use this on the dipole terms and the quadrupole terms: F =
[
E
( 0 ) ( x ') q +∑ i ∑ j
x i ∫ρ( x ) x (^) j d 3 x
∂ xi ' E (^) j ( 0 )
∑ i ∑ j ,k
x i ∫ρ( x ) x (^) j xk d 3 x
∂ xi '
∂ xk ' E (^) j ( 0 ) +...
] x '= 0
The electric fields do not depend on the unprimed variables and come out of the integrals, which was the point of the Taylor series expansion. After a little manipulation, we recognize the integrals that are left as the dipole moment and quadrupole moments: F =
[
E
( 0 ) ( x ') q +∑ i
x i
∂ xi ' p ⋅ E ( 0 )
∑ i
x i
∂ xi ' (∑ j ,k Q (^) j k
∂ xk ' E (^) j ( 0 ) +∑ j
∂ x (^) j ' E (^) j ( 0 ) ∫ρ( x^ ) r 2 d 3 x )+...
] x '= 0
Next note that the external field is created by charges outside our volume of interest, so: ∇⋅ E (^0 )= 0 or (^) ∑ i ∂ Ei ( 0 ) ∂ xi
This is the exact factor found in the last term shown, dropping that entire term out. Replacing the index notation with vector notation, we finally have: F = q E ( 0 ) ( 0 )+(∇ [ p ⋅ E ( 0 ) ]) 0 +
[
∑ j ,k Q (^) jk ∂ E (^) j ( 0 ) ∂ x (^) k ( x )
]) 0
Here we have switched primed variables to unprimed to match Jackson, and because the originally unprimed variables (the integration variables) are neatly tucked away now in the multipole moments. If we write the first term in terms of the potential, we can factor out the gradient operator: F =−∇
[
q Φ ( 0 ) − p ⋅ E ( 0 ) −
∑ j , k Q (^) jk ∂ E (^) j ( 0 ) ∂ xk ( x )+...
] 0
Let us compare this to the expansion (Jackson 4.24) of the energy W : W = q Φ( 0 )− p ⋅ E ( 0 )−
∑ i ∑ j Qij ∂ E (^) j ∂ xi
We see that using both equations, we recover the familiar expression: W =−∫ F ⋅ d x ( 0 )
