Using Gauss' Law to find the field at a point

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Homework Statement:
I have attached a pdf file of the entire problem. I am concerned with problem 23.03.
Relevant Equations:
##\epsilon_0 \oint \vec{E}\cdot d\vec{A} = q_{enc}##
Attached is problem 23.03 from Halliday and Resnick.

We have a sphere of uniform negative charge Q = -16e and radius R = 10cm. at the center of the sphere is a positively charged particle with charge q = +5e. We are supposed to use Gauss' law to find the magnitude of the electric field at point ##P_1## a distance r = 6cm from the center of the sphere. So we construct a sphere through the point ##P_1## and try to find the field. The outer charged sphere and the point charge both are symmetric, the field will always perpendicular to the sphere we constructed. Then we have

##\epsilon_0 E \oint dA = \epsilon_0 E (4\pi r^2) = q_{enc} \implies E = \cfrac{1}{4\pi \epsilon_0}\cfrac{q_{enc}}{r^2}##

But this is the same as if we just used coulomb's law to find the field of the single point charge (the one at the center) at ##P_1##. Why is that? The book doesn't explain this weird result. What happened to the field set up by the charged sphere?

It also doesn't tell us anything about the direction of E, just the magnitude. How can we know the direction of the field? Or was the direction necessary to set up the solution and I just didn't understand it?
 

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Answers and Replies

  • #2
etotheipi
Newton's shell theorem states that the electric field due to a charged shell, anywhere inside the shell, is zero. This result follows directly from application of Gauss' law to a spherical surface contained entirely inside the shell. So in your example, the electric field inside the shell is the sum of the electric field due to the inside point charge (Coulomb's law) and the electric field due to the shell (zero). So you can effectively 'ignore' the shell when calculating the field inside.

N.B. Coulomb's law can be derived from Gauss' law, and vice versa, when there is spherical symmetry.

It also doesn't tell us anything about the direction of E, just the magnitude. How can we know the direction of the field? Or was the direction necessary to set up the solution and I just didn't understand it?

The spherical symmetry dictates that the field has to be radial, i.e. it can only depend on the coordinate ##r##: ##\vec{E}(r) = E_r(r) \hat{r}##.

When you construct a spherical Gaussian surface, Gauss' law gives$$\frac{q}{\varepsilon_0} = \int_S \vec{E} \cdot d\vec{S} =\int_S E_r \hat{r} \cdot \hat{n} dS$$and since the surface is a sphere about the origin, ##\hat{r} \cdot \hat{n} = 1## and$$\frac{q}{\varepsilon_0} = \int_S E_r dS = 4\pi r^2 E_r$$Note that if ##q>0## then ##E_r > 0## (the field has a positive ##\hat{r}## component, and points outward, and if ##q<0## then ##E_r < 0## (the field has a negative ##\hat{r}## component, and points inward.
 
  • #3
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Newton's shell theorem states that the electric field due to a charged shell, anywhere inside the shell, is zero. This result follows directly from application of Gauss' law to a spherical surface contained entirely inside the shell. So in your example, the electric field inside the shell is the sum of the electric field due to the inside point charge (Coulomb's law) and the electric field due to the shell (zero). So you can effectively 'ignore' the shell when calculating the field inside.

Can you explain this more? How can the net field be zero anywhere inside the shell? That only makes sense if we are considering its center. If we have a uniform negatively charged sphere and we are considering a point closer to one side rather than the other, shouldn't the attraction from the closer side be stronger? The other side is too far away for it to completely cancel out. Is there any way to explain the shell theorem? I never understood it.
 
  • #4
vela
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Homework Statement:: I have attached a pdf file of the entire problem. I am concerned with problem 23.03.
Relevant Equations:: ##\epsilon_0 \oint \vec{E}\cdot d\vec{A} = q_{enc}##

But this is the same as if we just used coulomb's law to find the field of the single point charge (the one at the center) at P1. Why is that? The book doesn't explain this weird result. What happened to the field set up by the charged sphere?
##q_{\rm enc}## is the charge at the center plus the charge from the sphere that's inside the gaussian surface.
 
  • #5
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##q_{\rm enc}## is the charge at the center plus the charge from the sphere that's inside the gaussian surface.

We are looking at the first part of the problem, where ##P_1## is a point inside the charged sphere, so the charged sphere is external to the Gaussian surface. See figure 23-10 (a).
 
  • #6
vela
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We are looking at the first part of the problem, where ##P_1## is a point inside the charged sphere, so the charged sphere is external to the Gaussian surface. See figure 23-10 (a).
Ah, ok. I was going by your statement of the problem, which says a uniform sphere of charge, not a spherical shell of charge as stated in the book.
 
  • #7
etotheipi
Can you explain this more? How can the net field be zero anywhere inside the shell? That only makes sense if we are considering its center. If we have a uniform negatively charged sphere and we are considering a point closer to one side rather than the other, shouldn't the attraction from the closer side be stronger? The other side is too far away for it to completely cancel out. Is there any way to explain the shell theorem? I never understood it.

You need to take into account that there is more shell, i.e. more total charge, on the 'further' side than on the 'closer' side, assuming that the charge is uniformly distributed across the shell.

You should convince yourself that the shell theorem is true by summing the field due to consecutive rings of infinitesimal width via. calculus.
 

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