What Determines the Electric Field Inside a Spherical Conductor?

In summary, the magnitude of the electric field produced by the charge on the inner surface of a conducting spherical shell with a point charge at the center is q/4πε0r^2. The net electric field inside the conductor is zero due to the equal but opposite charges on the inner surface, but the point charge at the center creates an electric field according to answer D. Care must be taken when considering the inner charge, as it is not a point charge and may lead to incorrect calculations.
  • #1
pietastesgood
30
0

Homework Statement



1. Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer
radius R2. A point charge q is placed at the center of the cavity. The magnitude of the electric
field produced by the charge on the inner surface at a point in the interior of the conductor, a
distance r from the center, is:
A. 0
B. Q/4vπε0(R1)2
C. Q/4πε0(R2)2
D. q/4πε0r^2
E. Q/4πε0r^2

Answer is D.

Homework Equations



E=kq/r^2
flux=q/ε0

The Attempt at a Solution



I don't quite understand why there is an electric field within the conductor itself, which is what answer choice D is saying. Sure, there's a charge within the cavity, but since the net charge has to equal zero, there will be an equal but opposite charge on the inner cavity surface, which allows no flux to leave through a Gaussian surface within the conductor at a radius greater than the inner cavity.

Unless I'm actually just reading the problem wrong, and it's talking about just the charge on the inner surface, and ignoring the charge within the cavity. In that case, wouldn't the charge -q cause a field of magnitude q/4πε0(r-r1)^2? Any help would be appreciated!
 
Physics news on Phys.org
  • #2
pietastesgood said:

Homework Statement



1. Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer
radius R2. A point charge q is placed at the center of the cavity. The magnitude of the electric
field produced by the charge on the inner surface at a point in the interior of the conductor, a
distance r from the center, is:
A. 0
B. Q/4vπε0(R1)2
C. Q/4πε0(R2)2
D. q/4πε0r^2
E. Q/4πε0r^2

Answer is D.

Homework Equations



E=kq/r^2
flux=q/ε0

The Attempt at a Solution



I don't quite understand why there is an electric field within the conductor itself, which is what answer choice D is saying. Sure, there's a charge within the cavity, but since the net charge has to equal zero, there will be an equal but opposite charge on the inner cavity surface, which allows no flux to leave through a Gaussian surface within the conductor at a radius greater than the inner cavity.

Unless I'm actually just reading the problem wrong, and it's talking about just the charge on the inner surface, and ignoring the charge within the cavity. In that case, wouldn't the charge -q cause a field of magnitude q/4πε0(r-r1)^2? Any help would be appreciated!

Yeah, I think you have to read it carefully. There is no net electric field inside of the conductor. The charge at the center will create an electric field corresponding to answer D. The charges on the conductor will rearrange to cancel that. But the inner charge of -q is not a point charge. Be careful what field you think it makes.
 
  • #3
Oops, you're right. If you take the inner charge -q and use Gauss' law, you get EA=q/ε0, and A=4πr^2, so E=q/4πε0r^2. Thanks for the help!
 

FAQ: What Determines the Electric Field Inside a Spherical Conductor?

1. What is Gauss' Law?

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

2. What is the equation for Gauss' Law?

The equation for Gauss' Law is ∮E•dA = Qenclosed0, where ∮E•dA represents the electric flux through a closed surface, Qenclosed is the total electric charge enclosed by that surface, and ε0 is the permittivity of free space.

3. What is the significance of Gauss' Law?

Gauss' Law allows us to calculate the electric field at a point due to a distribution of charges, without having to consider the individual contributions from each charge. It also helps in understanding the symmetry and behavior of electric fields in different scenarios.

4. How is Gauss' Law applied in real-life situations?

Gauss' Law is applied in various real-life situations, such as calculating the electric field inside a capacitor, determining the electric field due to a charged metal sphere, or analyzing the electric field of a uniformly charged sheet. It is also used in the design and analysis of electrical systems and devices.

5. Can Gauss' Law be used for any type of charge distribution?

Yes, Gauss' Law can be used for any type of charge distribution, as long as the charge is enclosed by a closed surface. It is a general law and does not depend on the shape, size, or location of the charges.

Back
Top