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Atty. Santos' outline for ITL, DLSU Law
author suarez
O. PROBLEM STATEMENT
You have three charged spheres of radius “a”, one conducting (1), one having a uniform-charge-density uniform-charge-density smeared about its non-conducting volume (2), and one having a spherically-symmetric charge density that n varies radially as r (n > -3) -3) (2). Each has a total charge “Q”, making for a total charge ch arge of 3Q present. Use Gauss’s Theorem to obtain electric fields both inside and outside each sphere. I. MY PROOF
(1) Inside conducting sphere, field is zero, as we proved in 04 EM, p. 050, Pr 1.1b. (2) Inside uniformly-charged uniformly-charged sphere: field varies linearly in “r”. Specifically: E
(x)d x 3
0
d A
r
4 r 2 dr
Q 3 4 R 3
ˆ
2
0 4 r
[I.1]
r ˆ
Effecting this integration,[I.1], one gets: E
Q 4 0 R 3
[I.2]
r r ˆ
(3) Inside spherically-symmetrically-charged sphere: use same formula featured in [I.1]. To use[I.1]: we must construct charge-density function of the form (x ) Ar n , satisfying
( R ) dV
sphere
( 0)
0 and
Q . The later condition will require a bit of symbol-crunching: so here goes: Q
R
0
Ar n 4 r 2 dr 4 A n13 Rn 3 0n 3 n3
n43 ARn 3
[I.3]
no "ln" to worry about
Then: solve for “A” in [I.3] and get:
A
n3 4 Rn 3
Q
(x) Q
n 3 4 Rn 3
r n
[I.4]
So we‟re ready to use [I.4] in [I.1]:
E
( x)d x 3
4 0r 2
r
Q r 0
ˆ
n 3 4 R n 3
( r )n 4 ( r)2 dr 4 0 r 2
r
ˆ
Q Rnn33 r n 3 4 0 r 2
r ˆ
Q(n 3) r n 1 40
n 3 R
r ˆ
[I.5]
…for r < R. A glance at our ou r result, [I.5], with n = 0 (uniform charge density) in mind reproduces [I.4], which is a mite of reassurance that we did the integration correctly. Also: right at the surface: if we use n = 0 and r = R, we reproduce the result of Coulomb‟s law for an ordinary charge. Outside the three spheres, the field ought to be given by Coulomb‟s Law, straight up. Let‟s illustrate:
By superposition: the electric field at rme should be: E
Ei
i 3
Q
4 i 1 0
2 ri rme 1
[I.6]
II. BACKGROUND READING
III. QUESTIONS
Must I account for all three spheres‟ presence? We don‟t even have their position-vectors. Oh well, I made „em up… IV. THE ANSWER
The uniformly-charged sphere: compare [I.2] with: …you got it Now: a tougher one: compare [I.5] with:
Whoops…darn charge density….let‟s figure this out: E
theirs , ( r ) Ar n
0r
Q ( n 3)
n 1
0 (n 3)
r ˆ
4 a
n 3
r n1
0 (n 3)
r ˆ
Qr n 1 4 0a n3
r ˆ
There is a factor of “n+3” difference between mine and theirs. Better watch out… And: I‟ll make a plot if I have to. V. SCRAP BIN
