A question about electrostatics / Gauss's law

In summary, the problem involves a conducting spherical shell with a charge of 3.00 nC and a charge of -2 nC at the center. The electric field is to be found at different distances from the center. Using Gauss's law and the principle of electric flux, the equation (E)(r^2) = (Qinside)(ke) is derived. The solutions for parts (a) and (c) are obtained by plugging in the known charges. The solution for part (b) is 0, as charge within a conductor will lay at the surface. For part (d), since the electric field is 0 and the Gaussian surface within the conductor yields 0, the net charge within the conductor is
  • #1
physics213
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Homework Statement



Suppose a conducting spherical shell carries a charge of 3.00 nC and that a charge of -2 nC is at the center of the sphere. If the distance from the center to the inner shell is 2.00 m, and the distance from the center to the outer shell is 2.40 m, find the electric field at:

a.) r = 1.5 m from center
b.) r = 2.2 m from center
c.) r = 2.5 m from center
d.) What is the charge distribution on the sphere?

Homework Equations



Using Gauss's law and the principle of electric flux, I was able to correctly derive the equation:

(E)(r^2) = (Qinside)(ke)

E=electric field
ke = 9 x 10^9



The Attempt at a Solution



I was able to get parts (a) and (c) just be using the equation and plugging in the charges that I knew.

The answer for (b) = 0
the answer for (d) = 2.00 nC on the inner surface, 1.00 nC on the outer surface.


I have worked for like 2 days on this problem and couldn't figure it out. Please help! Thanks in advance.
 
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  • #2
Well the answer for b) is 0. It is a conductor and charge within a conductor will lay at the surface.

http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/gausur.html#c2

In d) since the field is 0, and the Gaussian surface within the conductor yields 0, then doesn't that mean that the net charge within is 0? If the center has -2nC, then won't the inner surface of the conductor necessarily have +2nC? and if 2 of them are on the inner surface ... what must be left of the 3nC that it is charged with to be lounging about on the outer surface?
 
  • #3
LowlyPion said:
Well the answer for b) is 0. It is a conductor and charge within a conductor will lay at the surface.

http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/gausur.html#c2

In d) since the field is 0, and the Gaussian surface within the conductor yields 0, then doesn't that mean that the net charge within is 0? If the center has -2nC, then won't the inner surface of the conductor necessarily have +2nC? and if 2 of them are on the inner surface ... what must be left of the 3nC that it is charged with to be lounging about on the outer surface?

I see. But why would 2 of them be on the inner surface?
 
  • #4
physics213 said:
I see. But why would 2 of them be on the inner surface?

What's at the center?

-2nC.

What will the closed Gaussian surface within the middle of the conductor be? 0.

If the net of the charge inside the Gaussian must be 0, then 2nC must reside on the inner surface.
 
  • #5
LowlyPion said:
What's at the center?

-2nC.

What will the closed Gaussian surface within the middle of the conductor be? 0.

If the net of the charge inside the Gaussian must be 0, then 2nC must reside on the inner surface.

ah ok. thanks a lot!
 

FAQ: A question about electrostatics / Gauss's law

What is electrostatics?

Electrostatics is the study of electric charges at rest. It deals with the behavior of stationary electric charges and the forces they exert on each other.

What is Gauss's law?

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

How is Gauss's law used in electrostatics?

Gauss's law is used to calculate the electric field at a point due to a specific charge distribution. It is also used to determine the net charge within a closed surface and to calculate the electric flux through a surface.

What is the importance of Gauss's law?

Gauss's law is important because it allows us to understand and predict the behavior of electric charges and electric fields. It also provides a mathematical tool for solving complex electrostatic problems with symmetry.

What are the limitations of Gauss's law?

Gauss's law is only applicable to static electric fields and cannot be used to analyze time-varying fields. It also assumes a vacuum or uniform medium, so it may not be accurate for materials with varying dielectric constants or conductivity.

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