Gauss Law Problem With A Spherical Conductive Shell

In summary, the conversation discusses a hollow metallic sphere with inner radius r1, outer radius r2, and a charge of magnitude Q located at a distance d<r1 from the centre. The electric field lines are drawn for different regions and Gauss's law is used to find the electric field. It is noted that the sphere is a conductor and the only place without an electric field is inside the shell. The electric field is calculated to be E=Q/(4πr^2ε0) for all locations outside the shell.
  • #1
Lancelot59
646
1
You are a hollow metallic sphere of inner radius r1, and outer radius r2. Inside is a charge of magnitude Q and a distance d<r1 from the centre.

First I need to draw the electric field lines for regions r<r1, r1<r<r2, and r2<r

Since the sphere is a conductor the only place where there is not an electric field is inside the shell. The point charge induces a charge on the conducting sphere, making it in turn create an electric field outside the sphere.

I then need to use Gauss's law to find the electric field where possible. I think this is correct:

[tex]\int \vec{E}\cdot d\vec{A}=\frac{Q_{enclosed}}{\epsilon_{0}}[/tex]
[tex]E\int d\vec{A}=\frac{Q}{\epsilon_{0}}[/tex]
[tex]E(4\pi r^{2})=\frac{Q}{\epsilon_{0}}[/tex]
[tex]E=\frac{Q}{4\pi r^{2}\epsilon_{0}}[/tex]

For all locations that are not inside the shell. Am I correct?
 
Last edited:
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  • #2
Looks fine to me.
 
  • #3
Thanks for the confirmation.
 

Related to Gauss Law Problem With A Spherical Conductive Shell

1. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric field at a point to the distribution of electric charge around it. It states that the electric flux through any closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space.

2. What is a spherical conductive shell?

A spherical conductive shell is a conductive material in the shape of a hollow sphere. It can be made from a variety of conductive materials, such as metal, and is often used in experiments and demonstrations to study electrostatics.

3. How do you apply Gauss' Law to a spherical conductive shell?

To apply Gauss' Law to a spherical conductive shell, we must first determine the charge enclosed by the surface of the shell. This can be done by calculating the total charge of the shell or by using the charge density if it is uniform. Then, we can use Gauss' Law to find the electric field at any point outside or inside the shell.

4. What is the electric field inside a spherical conductive shell?

The electric field inside a spherical conductive shell is zero. This is because the charges on the surface of the shell are distributed in such a way that the electric field they produce cancels out at any point inside the shell.

5. Can Gauss' Law be used to find the electric field outside a spherical conductive shell?

Yes, Gauss' Law can be used to find the electric field outside a spherical conductive shell. By drawing a Gaussian surface outside the shell and using the charge enclosed by that surface, we can apply Gauss' Law to calculate the electric field at any point outside the shell.

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