Electricity and Magnetism questions Gauss's law, groundingetc.

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Electricity and Magnetism questions....Gauss's law, grounding..etc.

Alright so I am finally starting to understand the material, but I have a few questions.

1. An insulating spherical shell with an inner radius of 0.1cm and outer radius of 0.3 cm carries a total charge of 20nC. Use Gauss's law to find an expression for the electric field at a distance of r=0.08cm, r=0.2cm, r=0.4cm.

So the area of a shell is 4πR^2. EA=Qin/E0, so E=Qin/(4πR^2E0).
Now I know for the r=0.4cm one I can just use Qin=20nC because the whole charge is enclosed, but for the one's where the whole charge isn't enclosed I believe I do Qin=[tex]\sigma[/tex]4πr^2, and so E=[tex]\sigma[/tex]4πr^2/(4πR^2E0)....is this equation correct? and I am a bit confused about what to make r and R and when...like when it is in the inner radius is R=0.1 cm or is it still 0.3cm and why if so? can someone help me out please?

2. An arc with a length of 6cm and a radius of 3cm carries a uniform charge of 10nC. Derive an expression for the magnitude and direction of electric field at the center.

Ok so I am not sure if you are suppose to use Gauss's law or Coulomb's law and integrate for this question...can someone help me get started and point me in the right direction please?

3. A hollow conducting sphere has a radii of 0.8m and 1.2m. It surrounds a charge of +300nC, and the hollow sphere carries a charge of -200nC. Find the charge density on its inner and outer surfaces. B) if the outer surface of the shell is grounded, then the charge on the inner surface is _________ and on the outer surface is _______.

Ok I already found the charge densities, they are -37.3 nC on the inner surface and 5.53 on the outer surface. However I am not sure what happens if the outer surface of the shell is grounded. :S
 

Answers and Replies

  • #2
LowlyPion
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For 1. you will want to remember that with Gaussian surfaces you are interested in the charge enclosed.

For 2. exploit the symmetry about the axis, such that you have charge elements dq that contribute twice the E at distance r times the cos of the angle from 0 to 1 radian. (Your arc is 2 radians, so just use half the arc and double in the direction of the axis. By symmetry the orthogonal components to the axis will cancel.)

For 3-b. after you attach it to ground, it looks like a capacitor to ground between the inner surface and the central charge. Hmmm. Didn't you already calculate that?
 
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1. Ok so for the r=0.0008m one EA=Qin/E0
A=4πr^2
Qin=[tex]\sigma[/tex]Ain=(20uc/4π0.003^2)(4π0.0008^2) (is 0.003 right or should I use 0.001 since 0.0008 is inside the 0.001cm radius...but the charge is from 0.1cm-0.3cm right so why would it matter?)
so Qin=1.42uC
and then E=Qin/(AE0)
so E=1.42uC/(4πr^2E0)
so since....ugh :S.....do I use 0.3cm as r for the area, or do I use 0.3cm-0.1cm=0.2cm....or is it 0.1 cm? and does it change from point to point or is the r for the area constant? This is what I am confused about....

3. So are you saying that the outer surface becomes 0 and the inner surface remains -37.3 nC?
 
  • #4
LowlyPion
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Since there is no charge inside .01cm what field is there at .008?

For 3. if the conductor is grounded, and the E = 0 inside the conductor, how else could there be no net charge inside a Gaussian spherical surface just farther away than the inside surface of the conductor?
 
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1.Ah I see, so the charge at 0.08cm is zero. Ok so then at 0.2 cm you would use the radius as 0.3 cm right?

3.If the gaussian surface is inside the area of the conductor then E=0....so the inside becomes zero and the outside becomes the outside plus the inside? sorry I am confused :S.
 
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And for 2. E=[tex]\int[/tex]KdQ/r^2
[tex]\lambda[/tex]=Q/l=10nC/6cm=166.7nC/m
r=0.03m
dQ=[tex]\lambda[/tex]
and we integrate from 0 radians to 1 radians....
Ok I am a little confused about how to set up the integral...cos is adjacent over hypotenuse, so adjacent I use as r or x...and then the hypotenuse.... :S
 
  • #7
LowlyPion
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1.Ah I see, so the charge at 0.08cm is zero. Ok so then at 0.2 cm you would use the radius as 0.3 cm right?

Not quite so fast. At .2 cm then you want to determine the net charge inside the sphere of radius .2cm. Figure that from the charge density times the volume of what is a spherical shell from .1 to .2 cm. Likewise then for calculating the field at .3 cm.
 
  • #8
LowlyPion
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And for 2. E=[tex]\int[/tex]KdQ/r^2
[tex]\lambda[/tex]=Q/l=10nC/6cm=166.7nC/m
r=0.03m
dQ=[tex]\lambda[/tex]
and we integrate from 0 radians to 1 radians....
Ok I am a little confused about how to set up the integral...cos is adjacent over hypotenuse, so adjacent I use as r or x...and then the hypotenuse.... :S

Make a drawing of your arc with a mid line intersecting that. Call that your x axis. By symmetry you can ignore the y components then because they are equal and oppositely positioned in y such that their components will cancel. (Half are pointing up and half down when you take your integral. You should satisfy yourself on this point and don't just take my word for it.)

Then that means that along your arc you have uniform dq's that are equally spaced at distance of the radius. That determines the |E| that each dq contributes to what you want to integrate. But as discussed above the "y" components are no longer of interest. Only the x components.

Now each r makes an angle θ with the x-axis, your line of symmetry. (Your hypotenuse of course being just the distance r.) And only the contribution of the |dE|, which along the x-axis is given by r*cosθ is what you are integrating. And for convenience since you can incorporate all of the dq's in the range of the arc by integrating over θ then you can express things as

dE = λ*r*cosθ dθ

where λ is your charge density. C/2

You can integrate then over the range -1 to 1 radian or 0 to 1 and double it, your choice.
 

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