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Electric field sphere calculations with a hollow cavity using Gauss' law

  • #1

Homework Statement


A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q. (picture attached)
a)What is the total surface charge q(int) on the interior surface of the conductor (i.e., on the wall of the cavity)?


Homework Equations


E=Kq/r2, K=constant
Gauss's Law: ∫= closed surface integral-
∫E⋅dA=Qenclosed0, ε0=constant

The Attempt at a Solution


a) Since the conductor is neutral this means it has just as many electrons as protons, however since it is a conductor the electrons are loosely bound and can move. These loosely bound electrons will migrate toward the outer edge of the hollowed portion of the sphere leaving an excess charge of -q on it. My intuition tells me this is correct, but how would I go about solving/proving this with gauss's law? is gauss's Law necessary?
If I used Gauss's Law would I draw my gaussian surface like (figure 1 attached)?
With the way I drew It I wouldn't know my radius.
So my next thought was to draw it with the same radius as the hollowed out portion, lets call it "r"
But what would my Q enclosed be? clearly I am only enclosing a (+)q. This is where I am stuck.
I know the answer is -q but cannot prove/explain/show why.

Thanks!
 

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Answers and Replies

  • #2
TSny
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If I used Gauss's Law would I draw my gaussian surface like (figure 1 attached)?
Yes, that's a good choice.
With the way I drew It I wouldn't know my radius.
You will not need to know the radius. But you will need to know an important fact about the electric field at points inside a conductor in electrostatic equilibrium.
 
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  • #3
Yes, that's a good choice.
You will not need to know the radius. But you will need to know an important fact about the electric field at points inside a conductor in electrostatic equilibrium.
Okay, I know that the electric field within a conductor in electrostatic equilibrium is zero (assume thats what you were talking about). However, if we had no charge in the cavity the electric field would still be zero within the conductor (since the problem told us it was neutral).
We know that in the cavity there is an electric field pointing radially away from the positive charge (q), when this electric field hits the edge of the metal sphere does it stop? does it keep going through? I realize that the force on loosely bound electrons is opposite that of a field (if you put a negative charge in a field, the force it experiences is opposite to that of the field), and I assume that electrons within a neutral conductor cannot experience an acceleration.

I know I have some components of my question answered but Im still not able to piece it all together.
 
  • #4
After thinking about it a bit more, this must be what is happening (Figure 2.jpg)?

Assuming this is correct, the electric field and force are going in opposite directions. Do they somehow negate each other?

Is the reason you don't need to know the radius of gaussian surface because:
∫E⋅dA=Qenclosed0→E=Qenclosed/(dAε0) Here Qenclosed=0, therefore the radius of gaussian surface doesn't matter?
 

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  • #5
Update on my knowledge: We know that the electric field inside a conductor is zero, therefore if we draw a gaussian surface like I did in figure 1.jpg (above) the total charge enclosed has to be zero to make the electric field=0. Therefore if we have +q inside the cavity, we must also have -q somewhere within our gaussian surface. Since negative charge is free to move, and since we introduced an electric field through the conductor, the negatively charged electrons will feel a force opposite that of the electric field, finally making them move on the wall of the cavity and have a magnitude of -q.
 
  • #6
TSny
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We know that the electric field inside a conductor is zero,
Yes.
therefore if we draw a gaussian surface like I did in figure 1.jpg (above) the total charge enclosed has to be zero to make the electric field=0.
It is not clear here why you can claim that the total charge enclosed must be zero in order to make the field zero inside the conductor. Can you justify that claim with the help of Gauss' law?
 
  • #7
Yes.
It is not clear here why you can claim that the total charge enclosed must be zero in order to make the field zero inside the conductor. Can you justify that claim with the help of Gauss' law?
How does this look?
∫E⋅dA=Qenclosed0→E(4πr2)=(+q)+(-q)/ε0→E=0
 
  • #8
TSny
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How does this look?
∫E⋅dA=Qenclosed0→E(4πr2)=(+q)+(-q)/ε0→E=0
It seems to me that you have things turned around a little. You already know that E = 0 at points inside the conductor. You want to use that to construct an argument for why the induced charge on the inner surface of the cavity must be -q.

For the gaussian surface that you chose in your figure 1, what can you say about the total electric flux through the surface and why?
 
  • #9
It seems to me that you have things turned around a little. You already know that E = 0 at points inside the conductor. You want to use that to construct an argument for why the induced charge on the inner surface of the cavity must be -q.

For the gaussian surface that you chose in your figure 1, what can you say about the total electric flux through the surface and why?
okay, so we know E=0, therefore the only way to make E=0 is if the charge on the wall of the cavity is -q. The only possible way to make the electric field zero is if the enclosed charge sums to zero?

The total electric flux though the surface would be constant? why? because of gauss's law? probably not...
not sure how to answer that, but would love to know why so i can really grasp all the aspects of this.
Thanks!
 
  • #10
TSny
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okay, so we know E=0, therefore the only way to make E=0 is if the charge on the wall of the cavity is -q.
This is what you want to show with a logical argument. That is, show E = 0 implies that the induced charge on the surface of the cavity is -q.

You have constructed a good gaussian surface in which all points on the surface are inside the conducting material. You also know that E = 0 at all points insided the conducting material. What can you say about the value of the surface integral ∫EdA for the gaussian surface that you chose? (You don't need to use Gauss' law to answer this particular question.)
 
  • #11
This is what you want to show with a logical argument. That is, show E = 0 implies that the induced charge on the surface of the cavity is -q.

You have constructed a good gaussian surface in which all points on the surface are inside the conducting material. You also know that E = 0 at all points insided the conducting material. What can you say about the value of the surface integral ∫EdA for the gaussian surface that you chose? (You don't need to use Gauss' law to answer this particular question.)
We could say that the value of the surface integral is zero. Since the electric field within the conductor (also where i drew my gaussian surface) is zero we can say that ∫E⋅dA→0A→0
 
  • #12
TSny
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We could say that the value of the surface integral is zero. Since the electric field within the conductor (also where i drew my gaussian surface) is zero we can say that ∫E⋅dA→0A→0
Yes. Very good. Knowing this, what can you conclude from Gauss' law?
 
  • #13
That the electric field within a conductor in electrostatic equilibrium=0, no matter how much charge we place in the cavity that statement holds.
 
  • #14
TSny
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That the electric field within a conductor in electrostatic equilibrium=0, no matter how much charge we place in the cavity that statement holds.
No, we already knew that E = 0 at any point inside any conducting material in electrostatic equilibrium. You used that to conclude that ##\int \vec E \cdot \vec{dA} = 0## for your gaussian surface. Now, think about the statement of Gauss' law. When written as an equation, the left side of Gauss' law is ##\int \vec E \cdot \vec{dA}##. What's on the right side of Gauss' law?
 
  • #15
No, we already knew that E = 0 at any point inside any conducting material in electrostatic equilibrium. You used that to conclude that ##\int \vec E \cdot \vec{dA} = 0## for your gaussian surface. Now, think about the statement of Gauss' law. When written as an equation, the left side of Gauss' law is ##\int \vec E \cdot \vec{dA}##. What's on the right side of Gauss' law?
The charge enclosed over a constant
 
  • #16
No, we already knew that E = 0 at any point inside any conducting material in electrostatic equilibrium. You used that to conclude that ##\int \vec E \cdot \vec{dA} = 0## for your gaussian surface. Now, think about the statement of Gauss' law. When written as an equation, the left side of Gauss' law is ##\int \vec E \cdot \vec{dA}##. What's on the right side of Gauss' law?
so the charge enclosed has to equal zero to make the statement true 0=0
 
  • #17
TSny
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so the charge enclosed has to equal zero to make the statement true 0=0
Good. Almost home.
 
  • #18
Good. Almost home.
So knowing that within a conducting sphere in electrostatic equilibrium the field equals 0, we found that the total charge within has to also be zero?
 
  • #19
TSny
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So knowing that within a conducting sphere in electrostatic equilibrium the field equals 0, we found that the total charge within has to also be zero?
Yes, the total charge enclosed by your gaussian surface must be zero. Can you use this to deduce how much charge is on the surface of the cavity?
 
  • #20
Yes, the total charge enclosed by your gaussian surface must be zero. Can you use this to deduce how much charge is on the surface of the cavity?
Ahh yes, so we know that ∫E⋅dA=0 because E=0 within a conductor, therefore 0=Qenclosed0, we know that we have +q within the cavity so when solving the below equation for x we see that x=-q
0=(+q)+x/ε0
 
  • #21
TSny
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Ahh yes, so we know that ∫E⋅dA=0 because E=0 within a conductor, therefore 0=Qenclosed0, we know that we have +q within the cavity so when solving the below equation for x we see that x=-q
0=(+q)+x/ε0
That's it!

It would be better to arrange your parentheses on the right as 0=(+q+x)0
 
  • #22
That's it!

It would be better to arrange your parentheses on the right as 0=(+q+x)0
Okay! Thank you so much! I really appreciate the time you've spent!!! :)
 

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