EE381 Solutions of HW5 & Practice Problems · 2019. 3. 30. · r = 3 m contains a unifolm charge...
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EE381 Solutions of HW5 & Practice Problems Solution (d)
Transcript of EE381 Solutions of HW5 & Practice Problems · 2019. 3. 30. · r = 3 m contains a unifolm charge...
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EE381
Solutions of HW5 & Practice Problems
Solution (d)
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Solution:
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Solution
Notes: (i) R in Spherical coordinates is always positive.
(ii) There is a singularity at R = 0. The integral from z
= -4 m to +2 m needs to be carried out in 2 separate
integrals.
Since the field is in the radial direction described by Spherical
coordinates (spherical symmetry), the potential energy change
from z = -4 to 0 is exactly the same as that from z = 4 to 0.
To make it less confusing, the following integral is performed to
avoid using negative z:
V5.41818
ˆ18
ˆˆ18
ˆ
2
0
0
4
2
0 2
0
4 2
=
−−
−−=
⋅+⋅−= ∫∫
RR
dRR
dRR
VAB
zzzz
This is the same as moving from z = 4 to 2m:
V5.418
ˆ18
ˆ
2
4
2
4 2=
−−=⋅−= ∫
RdR
RV
ABzz
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