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Homework 2
4-2.A sphere of radius a has its center at the origin and a charge density given by p = Ar2 A =
constant. Another of radius is concentric with the first. Find the flux cj E. dii through the
surface of the larger sphere. From Gauss' Law,
Because spheres are concentric, The total Qin of smaller can be by 21r (^) 1r Qin = 1 p(r') dr' r r fa p(r')(r')2 sin B' dr' dB' dq/ V(a) )0 )0 0 (1r r
1r
fa A(r')2(r')2 dr' dB' dq/ A f21r J1r ~ (^) sin B' dB' dq/
)0 Jo 0 0 0 5
EO
contains
21r 2Aa
(^5) J 1+1 5 o hl'>r,,,,tnlrl'> the flux through (^) surface of larger
Qin
is
by an infinitely long cylinder of axis coincides with the line of the cylinder carries a of constant density cr. Find E everywhere. What particular value of cr will make E = 0 for all the charged cylinder? Is your answer
Due to both the cylinder and line length,^ E cannot depend on^ our^ I",..':of-ir....^ on^ z-axis
Due to symmetry of the line & E cannot on our <p in
must be (^) P component. E = Ep(P)fi.
- Case 1 : P < Po The cylinder around the line charge behaves similar to a conductor where the charge lies on the surface.
Inside any conductor, the electric field from surface charge is zero, thus ECYlinder = 0 in this case.
Thus, this case simply becomes the scenario a line charge where
A " (4 - 11)
2rcEopP,
= f/p(p)p, ftc +
Qin
L
Ep (P)2rcpL
If E = 0 when Po < P, then
A+
Which makes sense, as 2rcPo(J -A, providing equal and opposite to the line
for any desired length.
.. What is the smallest value of z which the potential due to the disc can be calculated as if it were a point without an error greater than 1 percent? For a point potential is calculated by 4>(T) = Q, (S The charge Q the circular disc is calculated by Q = rra^2 q Thus, treating the disc as a pOint charge z axis, potential is calculated by rra^2 q
4>(z) = =
In to have accuracy desired,
-0.01 <
a^2 q
< 0.01 => -0.01 < 1 - q^ 41zlEo < 0.
2100 [(a^
(^2) + Z2)1/2 -Izl]
0.99 < 2 [( az 2 2 + < 1.01 => < -z 2 <
a^2 a^2 a^2 2.02 <^ +^ Z^ 4)1/2^ _^ <^ 1.98 ->..-I'^ 2.02 +^ < (^ a^2 z 2 +^ <^ 1.98+ a^4 a^2 z^2 a^2 z 2 4.0804 + 1.01 + Z4 < + Z4 < 3.9204 + 0.99 +
4.0804 -101 < 0 <--+ 99 a^2 101 > 4.0804 => Izi >^ --a;:::::^ ..J25a^ =^ Sa
Therefore, in order to calculate the potential as a point charge without an error^ than 1^ no,',..o.,1"
z must be than;::::: Sa.
