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July 3, 2015

July 3, 2015

Posted by **Bobby** on Sunday, March 23, 2008 at 6:29pm.

1. A solid metal sphere of radius R has charge +2Q. A hollow spherical shell of radius 3R placed concentric with the first sphere has net charge -Q.

a) Describe the electric field lines both inside and outside the spheres (as if you had to draw field lines on the chart-no calculations).

b) Use Gauss' law to find an expression for the magnitude of the electric field between the spheres at a distance r from the center of the inner sphere (R<r<3R)

c) Calculate the potential difference between the two spheres.

d) The spheres are then connected by a conducting wire for long enough that the charges can redistribute. What is the final distribution charge on the two spheres?

e) The wire connecting the spheres is removed and the outside sphere is grounded. What is the charge distribution on the two spheres?

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?

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.

c) Set up an integral to evaluate the potential at the surface of hte cylinder (dont solve).

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?

e) What is the effect (qualitative) on the potential at the surface of the insulator.

- Physics-Gauss' Law, hard -
**Damon**, Sunday, March 23, 2008 at 6:55pma)

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

b)

R<r<3R

E = 2 Q/(4 pi r^2 eo)

c)

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.

- Physics-Gauss' Law, hard -
**Bobby**, Sunday, March 23, 2008 at 7:06pmDamon, 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 =].

- Physics-Gauss' Law, hard -
- Physics-Gauss' Law, hard -
**Damon**, Sunday, March 23, 2008 at 7:26pm2. 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.

E

inside

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

outside

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.

- Physics-Gauss' Law, hard -
**Anonymous**, Tuesday, March 25, 2008 at 9:21pmThank you so much Damon!

- Physics-Gauss' Law, hard -
**Musa**, Tuesday, June 30, 2009 at 6:55pme,