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Jackson 2.23 Homework Problem Solution
Dr. Christopher S. Baird University of Massachusetts Lowell PROBLEM: A hollow cube has conducting walls defined by six planes x = 0, y = 0, z = 0, and x = a , y = a , z = a. The walls z = 0 and z = a are held at constant potential V. The other four sides are at zero potential. (a) Find the potential Φ ( x , y , z ) at any point inside the cube. (b) Evaluate the potential at the center of the cube numerically, accurate to three significant figures. How many terms in the series is it necessary to keep in order to attain this accuracy? Compare your numerical result with the average value of the potential on the walls. See Problem 2.28. (c) Find the surface-charge density on the surface z = a. SOLUTION: The problem contains no charge, so the electric potential is described everywhere inside the cube by the Laplace equation: ∇ 2 = 0 This problem has a boundary, the cube, that best matches rectangular coordinates. In rectangular coordinates, the Laplace equation becomes: ∂ 2 ∂ x
2 ^
2 ∂ y
2 ^
2 ∂ z
2 =^0
Using the method of separation of variables, the most general solution to this equation is: x , y , z = A 0 B 0 x A 0 B 0 y A 0 B 0 z ∑ A 0 B 0 x A e i y B e − i y A e z B e − z ∑ A e i x B e − i x A 0 B 0 y A e z B e − z ∑ , A e i x B e − i x A e i y B e − i y A e z B e − z Apply the boundary condition, Φ ( x = 0, y , z ) = 0
0 = A 0 A 0 B 0 y A 0 B 0 z ∑ A 0 A e i y B e − i y A e z B e − z ∑ A B A 0 B 0 y A e z B e − z ∑ , A B A e i y B e − i y A e z B e − z This must be true for all y and z , so that each term must vanish separately. This forces Aα 0 = 0 and Bα = - Aα. The solution now becomes: x , y , z = B 0 x A 0 B 0 y A 0 B 0 z ∑ B 0 x A e i y B e − i y A e z B e − z ∑ A sin x A 0 B 0 y A e z B e − z ∑ , A sin x A e i y B e − i y A e z B e − z Similarly, the boundary condition, Φ ( x , y = 0, z ) = 0 leads to Aβ 0 = 0 and Bβ = - Aβ which gives the solution: x , y , z = B 0 x B 0 y A 0 B 0 z ∑ B 0 x A sin y A e z B e − z ∑ A sin x B 0 y A e z B e − z ∑ , A sin x A sin y A e z B e − z Now apply the boundary condition Φ ( x = a , y , z ) = 0 0 = B 0 a B 0 y A 0 B 0 z ∑ B 0 a A sin y A e z B e − z ∑ A sin a B 0 y A e z B e − z ∑ , A sin a A sin y A e z B e − z The only way this can be true for all y and z is if Bα 0 = 0 and α = n π/ a where n = 0, 1, 2,... x , y , z =∑ n An sin n x a B 0 y A e z B e − z ∑ n , An sin n x a ^ A sin y A e z B e − z Similarly, apply the boundary condition Φ ( x , y = a , z ) = 0 to get Bβ 0 = 0 and β = m π/ a where n = 0, 1, 2,... x , y , z =∑ n , m An sin n x a ^ Am sin m y a A e ^ z B e −^ z^
An , m =
8 V
n m 2 e −^12 n^2 m^2 ^2 cosh / 2 (^) n 2 m 2 Bn , m =
8 V
n m 2 e 1 2 n (^2) m (^2) 2 cosh / 2 (^) n 2 m 2 The final solution is now: x , y , z = (^) ∑ n ,m odd
sin
n x
a
sin
m y
a ^
8 V
n m
e −^12 n^2 m^2 ^2 ez^ ^ n (^2) m (^2) (^2) / a 2 e 1 2 ^ n (^2) m (^2) 2 e − z^ n^ (^2) m (^2) (^2) / a 2 cosh / 2 (^) n 2 m 2
x , y , z = (^) ∑ n ,m odd
16 V
n m
2 sin
n x
a
sin
m y
a
cosh / 2 (^) n 2 m 2 2 z / a − 1 cosh / 2 (^) n 2 m 2
(b) Evaluate the potential at the center of the cube numerically, accurate to three significant figures. How many terms in the series is it necessary to keep in order to attain this accuracy? Compare your numerical result with the average value of the potential on the walls. See Problem 2.28. The potential at the center of the cube is: a /2, a / 2, a / 2 =
16 V
2 ∑ n ,m odd
n m
sin
n
sin
m
cosh / 2 n 2 m 2
Φ( a /2, a /2, a / 2 )= V [
π
2 sin(
π
sin
π
2 )(^
cosh ((π/ 2 ) √ 2 ) )
2
sin
sin
2 ^
cosh / 2 10
2
sin
sin
2 ^
cosh / 2 10
2
sin
sin
2 ^
cosh / 2 26
... ]
a /2, a / 2, a / 2 = V [0.3476−0.0075−0.00750.0002...] Here is solution, to three significant figures, if you only keep the first term, or the first two terms, etc: 1 term a /2, a /2, a / 2 = V [0.348 ] 2 terms a /2, a /2, a / 2 = V [0.340] 3 terms a / 2, a / 2, a / 2 = V [0.333] 4 terms a /2, a /2, a / 2 = V [0.333]
We only have to keep the first three terms to have the answer accurate to three significant figures. It is obvious that the solution is converging to: a / 2, a /2, a / 2 =
V
There are six walls on the cube and two sides have a non-zero potential V , so the average value of the potential on the sides of the cube is Φave on surf =^ 2 V 6 =^ V
- This leads to the interesting conclusion that: a / 2, a /2, a / 2 =ave on surf (c) Find the surface-charge density on the surface z = a. = [
d dn ] n = a We have solved the potential on the inside of the cube, so we can only use that potential to find the surface charge density on the inside of the z = a surface. The normal to the inside surface is in the negative z direction so that: = [
d dz ] z = a = [
d dz ∑ n , m odd
16 V
n m 2 sin^ n x a sin m y a cosh / 2 n 2 m 2 2 z / a − 1 cosh / 2 n 2 m 2 ] z = a = [ 0 ∑ n ,m odd
16 V
n m 2 sin n x a sin m y a sinh / 2 n 2 m 2 2 z / a − 1 cosh / 2 n 2 m 2 / a (^) n 2 m 2 ] z = a σ= 16 ϵ 0 V π a ∑ n , m odd √ n 2
- m 2 n m sin ( n π x a ) sin ( m π y a ) (tanh^ ((π/^2 )^ √ n (^2) + m (^2) ) )
