Insulating sphere

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Homework Statement


An insulating sphere with a radius of 0.100 m has a charge of 0.965 nC uniformly distributed throughout its volume. The center of the sphere is a distance 0.295 m above a large uniform sheet that has a charge density of -7.90 nC.

Find the distance from the center of the sphere to the point inside the sphere at which the electric field is zero.


Homework Equations


r < R

E = 1/4pi*E_o * Qr/R^3


Is there only 1 sphere that is directly above the sheet that is 0.295m ? I'm not exactly sure how the picture looks, so I wanted to verify this and then what to do next.
 

Answers and Replies

  • #2
Doc Al
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Is there only 1 sphere that is directly above the sheet that is 0.295m ?
Yes. Picture the large sheet as being in the x-y plane at z = 0. The sphere of charge can be pictured as being centered at point (0, 0, 0.295).
 
  • #3
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Okay so having that out of the way we have the following givens:

r = 0.100m
+Q = 0.965 nC
R = 0.295
-Q = -7.90

The field lines out of the sphere is going outward since its a positive charge.

We want to find the distance where the electric field is zero.

How should I start this problem based on a Solid insulating sphere with radius R and charge Q distributed uniformly throughout the volume?

Since the question asks inside the sphere it should be the case that:

r < R

E = 1/4pi*E_o * Qr/R^3

We want E = 0 ?

1/4*pi*E_o * Qr/R^3 = 0 ?
 
  • #4
Doc Al
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Since the question asks inside the sphere it should be the case that:

r < R

E = 1/4pi*E_o * Qr/R^3
Good. That's the field from the uniform sphere of charge.

We want E = 0 ?

1/4*pi*E_o * Qr/R^3 = 0 ?
No. You want the point where the total electric field due to all the charge is zero, not just the field from the sphere of charge.

The field at any point is the sum of the field from the plane plus the field from the sphere.
 
  • #5
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Good. That's the field from the uniform sphere of charge.


No. You want the point where the total electric field due to all the charge is zero, not just the field from the sphere of charge.

The field at any point is the sum of the field from the plane plus the field from the sphere.
I currently have the following equation:

(-7.90 / 2*E_0) + (1/4*pi*E_0) = .965/r^2

I sum up the charge of the sheet with the sphere and solve for r I get r = 0.45 but thats incorrect and I was wondering if I was attempting this correctly?
 
  • #6
Doc Al
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I currently have the following equation:

(-7.90 / 2*E_0) + (1/4*pi*E_0) = .965/r^2
I don't understand this equation. You should have something that looks like: Field from sheet + Field from sphere = 0.

Also, you should be able to point to the spot on your diagram (assuming you've made one) where this condition would be met. That will help you visualize what's going on.
 
  • #7
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I don't understand this equation. You should have something that looks like: Field from sheet + Field from sphere = 0.

Also, you should be able to point to the spot on your diagram (assuming you've made one) where this condition would be met. That will help you visualize what's going on.
Yes that is what that equation represents field from sheet + field from sphere = 0. I know the distance should be above the center point of the sphere.
 
  • #8
Doc Al
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Yes that is what that equation represents field from sheet + field from sphere = 0.
I'm not seeing it. Your first term looks like the field from the sheet. But what's the second and third terms? And why isn't the sum set equal to zero?

I know the distance should be above the center point of the sphere.
Right. (Just checking.)
 
  • #9
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Is this the electric field for a sphere = 1/4pi*E_o * Qr/R^3
 
  • #10
Doc Al
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Is this the electric field for a sphere = 1/4pi*E_o * Qr/R^3
Yes, as I thought you had stated several posts ago.
 
  • #11
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okay I'm a little confused on which "r" we're looking for, here's my math so far:

[tex] \frac {\sigma} {2*e_o} = \frac {1} {4*pi*e_o} * \frac {Qr} {R^{3}} [/tex]

= [tex] \frac {-7.9} {2*(8.85*10^{-12})} = \frac {1} {4*pi*(8.85*10^{-12})} * \frac {.965*r} {.295^{3}} [/tex]

Is big R = 0.295 the distance from center of sphere to the sheet ?

= [tex] 3.49*10^{-11} = \frac {6.70*10^{-12}*r} {.025672} [/tex]

= [tex] \frac {3.49*10^{-11}} {6.70*10^{-12}} * (.025672) = r => 1.3*10^{-25} [/tex]

and I don't think that is correct =/
 
Last edited:
  • #12
Doc Al
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okay I'm a little confused on which "r" we're looking for, here's my math so far:

[tex] \frac {\sigma} {2*e_o} = \frac {1} {4*pi*e_o} * \frac {Qr} {R^{3}} [/tex]

= [tex] \frac {-7.9} {2*(8.85*10^{-12})} = \frac {1} {4*pi*(8.85*10^{-12})} * \frac {.965*r} {.295^{3}} [/tex]

Is big R = 0.295 the distance from center of sphere to the sheet ?
No, big R is the radius of the sphere. (The formula for the field within the uniform sphere of charge has nothing to do with the distance to the sheet.)

Also: Careful with your arithmetic. Simplify as much as possible before plugging in numbers. (For example, the [tex]\epsilon_0[/tex] cancels from both sides.)
 
  • #13
I think it's easier to do this problem using Gauss's law to simplify spherical E in terms of charge density. Then set that equal to Eplate and after some simple algebra you have your answer.
 
  • #14
Doc Al
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I think it's easier to do this problem using Gauss's law to simplify spherical E in terms of charge density.
That part's already solved. :wink:
Then set that equal to Eplate and after some simple algebra you have your answer.
That's exactly what he's doing.
 
  • #15
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That part's already solved. :wink:

That's exactly what he's doing.
Okay let's try again from the start:

[tex] \frac {\sigma} {2*e_o} = \frac {1} {4*pi*e_o} * \frac {Qr} {R^{3}} [/tex]

= [tex] R^3*\frac {\frac {\sigma} {2*e_o}} {\frac {1} {4*pi*e_o}} [/tex]

= [tex] R^3*\frac {\frac {\sigma} {\frac {1} {2*pi}}} {Q} [/tex]

Is this correct, I think I might have done something wrong, but it's pretty late. Thank you
 
Last edited:
  • #16
Doc Al
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Yes, that's correct. Rewrite it more simply as:
[tex]r = \frac{2 \pi \sigma R^3}{Q}[/tex]
 
  • #17
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thank you.
 

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