Posted by Bobby on Sunday, March 23, 2008 at 6:29pm.
inside the inner sphere, a closed spherical surface encloses no charge, so E is 0
between spheres, the outer sphere is irrelevant, only the 2 Q matters
E drops off with 1/r^2, the surface area of spherical surface.
outside sphere 2
we surround a total of 2Q-Q = 1 Q
field again drops off as 1/r^2
E = 2 Q/(4 pi r^2 eo)
integral E dr = [Q/(2 pi eo)] dr/r^2
= [Q/(2 pi eo)] [-1/r] at 3R - at 1R
= [Q/(2 pi eo)] [ -1/3R +1/R ]
= (2/3)[Q/(2 pi eo R)]
= Q/(3 pi eo R)
d) the E between spheres must be zero or there would be a potential difference between spheres, so all charges must move to the outer sphere, net 1 Q
e) all gone.
Damon, you are amazing...
Tell me if you get the second one please, I'm still having trouble.
In the meantime I temporarily took over your role answering all the math questions =].
2. A long, insulating cylinder has a uniform charge density, ro=-2*mu*C/m^3, and a radius, R=15cm.
a) What is the total charge on a 40cm length of that cylinder?
volume of .4 m length = pi r^2 L = pi (.15)^2 (.4) = .02827 m^3
Q = ro * volume = -.05655 *10^-6
= -5.66*10^-8 C
b) Use Gauss' Law to find the electric field both inside and outside of the cylinder as a function of the distance from the center of the cylinder, r.
Q enclosed = ro pi r^2 L
area of Gauss surface = 2 pi r L
E = Qenclosed/area eo = ro pi r^2 L/2 pi r L eo
= ro r/2 eo
E = Q at R length L / 2 pi r eo L
= ro pi R^2 L / 2 pi r eo L
= ro (R^2/r)/2 eo
E *area = Qenclosed / 4 pi r^2 eo
c) Set up an integral to evaluate the potential at the surface of hte cylinder (dont solve).
well, I suppose we call it 0 at the center so
integral E dr = integral inside
= (ro /2 eo) r dr
An isolated (ungrounded) conducting shell of inside radius 30cm and outside radius 40cm is now snapped together around the insulating cylinder.
d) What is the effect (qualitative) on the electric field at r=50cm?
still surround the same old charge, -Q moves inside, + Q moves to outer surface. No effect
e) What is the effect (qualitative) on the potential at the surface of the insulator. No effect
Now if you ground that conducting shell, that kind of defines your zero of potential there, whereas I defined it at the center of the problem. That would change Voltages but not voltage differences.
Thank you so much Damon!
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