# Using Gauss' Law to find the field at a point

• rtareen
In summary: I'm not sure what you're saying. The charged sphere is external to the surface of the Gaussian? Are you saying that the electric field is the same no matter where on the surface you are?
rtareen
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?

#### Attachments

• Problem23.03.pdf
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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.

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

rtareen said:
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.

etotheipi said:
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.

rtareen said:
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.

vela said:
##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).

rtareen said:
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.

rtareen said:
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.

## 1. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric field at a point to the charge enclosed by a closed surface surrounding that point.

## 2. How do you use Gauss' Law to find the electric field at a point?

To use Gauss' Law, you must first choose a closed surface that encloses the point where you want to find the electric field. Then, you calculate the total charge enclosed by that surface and use it in the equation: E = Q/ε₀A, where E is the electric field, Q is the total charge enclosed, ε₀ is the permittivity of free space, and A is the area of the closed surface.

## 3. Can Gauss' Law be used for any shape of closed surface?

Yes, Gauss' Law can be used for any shape of closed surface as long as it completely encloses the point where you want to find the electric field.

## 4. What is the significance of Gauss' Law in electromagnetism?

Gauss' Law is significant because it provides a simple and elegant way to calculate the electric field at a point due to a distribution of charges. It also helps in understanding the symmetry of electric fields and their relationship to the charges that create them.

## 5. Are there any limitations to using Gauss' Law?

Yes, there are some limitations to using Gauss' Law. It can only be applied in situations where the electric field is constant over the chosen closed surface. It also assumes that the charge distribution is static and does not change over time.

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