Gauss's Law and Electric Field Question.

In summary, the problem involves a plastic, spherical shell with uniform charge q=16e and radius R, with a particle with charge Q = 5e at the center. The question asks for the electric field (magnitude and direction) at point P1 at a radial distance a from Q. The solution can be found using Gauss' Law, which relates the electric field at points on a closed surface to the net charge enclosed by that surface. In this case, the electric field at P1 is only due to Q, as the Gaussian sphere that goes through P1 only encloses Q and not q. Therefore, the electric field due to q is zero at P1. This is confirmed by the shell theorem, which states that the
  • #1
sanhuy
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2

Homework Statement


figure shows, in cross section, a plastic, spherical shell
with uniform charge q=16e and radius R. A particle
with charge Q = 5e is at the center.What is the electric field
(magnitude and direction) at (a) point P1 at radial distance a from Q

My question is at the point P1 is the electric field that the book found using gauss's law the total electric field or just the electric field due to Q (the point charge only).

This is worked out example 23.03 in the book "fundementals of physics 10e" by halliday and resnick.
Thanks.

Homework Equations


Q/ε = ∫Eda

The Attempt at a Solution


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Last edited:
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  • #2
sanhuy said:

Homework Statement


figure shows, in cross section, a plastic, spherical shell
with uniform charge q=16e and radius R. A particle
with charge Q = 5e is at the center.What is the electric field
(magnitude and direction) at (a) point P1 at radial distance a from Q

My question is at the point P1 is the magnitude of the electric field found using guass's law due to Q, or both Q and q?

What does Gauss' Law state?
 
  • #3
ehild said:
What does Gauss' Law state?
I believe it states that the total electric flux out of / into a closed surface is equal to the charge enclosed divided by the permittivity. it pretty much relates the electric field at points on a closed surface to the
net charge enclosed by that surface.
 
  • #4
Sorry, Let me rephrase my question because there may be some misinterpretation. is the electric field that the book found using gauss's law the total electric field or just the electric field due to Q (the point charge only).
 
  • #5
Are you familiar with the shell theorem?
 
  • #6
Orodruin said:
Are you familiar with the shell theorem?
No, not really. :\
 
  • #7
wait, is the electric field inside of a shell zero due to symmetry if the charge is uniformly distributed on the spherical shell?
 
  • #8
sanhuy said:
Sorry, Let me rephrase my question because there may be some misinterpretation. is the electric field that the book found using gauss's law the total electric field or just the electric field due to Q (the point charge only).
E is the electric field at the points of the Gaussian surface (the red sphere in your figure).
What charge is enclosed by your red Gaussian sphere that goes through P1? Is q inside it?
 
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  • #9
ehild said:
E is the electric field at the points of the Gaussian surface (the red sphere in your figure).
What charge is enclosed by your red Gaussian sphere that goes through P1? Is q inside it?

Well it only encloses the charge Q not q. So is the electric field calculated at P1 only due to Q? and the electric field due to q is zero at point P1 because of gauss's law?

is this correct?
thanks.
 
  • #10
sanhuy said:
Well it only encloses the charge Q not q. So is the electric field calculated at P1 only due to Q? and the electric field due to q is zero at point P1 because of gauss's law?

is this correct?
thanks.
Right.
 
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  • #11
Thanks guys.
 

1. What is Gauss's Law?

Gauss's Law is a fundamental law in electrostatics that relates the electric flux through a closed surface to the charge enclosed by that surface. It states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space.

2. How is Gauss's Law used to calculate the electric field?

Gauss's Law can be used to calculate the electric field at a point due to a charge distribution by using a closed surface that encloses the charge. The electric field is then calculated by taking the flux through that surface and dividing it by the permittivity of free space.

3. What is the significance of Gauss's Law in electrostatics?

Gauss's Law is significant because it provides a powerful tool for calculating electric fields and understanding the behavior of charges. It also helps to simplify complex charge distributions and can be used to derive other important laws in electrostatics.

4. Can Gauss's Law be applied to all charge distributions?

Yes, Gauss's Law can be applied to all charge distributions, as long as the charge is enclosed by the chosen surface and the electric field is constant over that surface. However, it is more useful for symmetrical charge distributions.

5. How does Gauss's Law relate to Coulomb's Law?

Gauss's Law and Coulomb's Law are closely related. Coulomb's Law describes the force between two point charges, while Gauss's Law relates the electric field to the charge distribution. Coulomb's Law can be derived from Gauss's Law for a point charge, and Gauss's Law can be used to calculate the electric field due to a continuous charge distribution, which is more difficult to do with Coulomb's Law.

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