# Gauss' Law question about a conducting rod

• Physicslearner500039
In summary, the conversation discusses the calculation of electric field and net charge in different scenarios involving a solid sphere and conducting shells. The main point is that the electric field inside a conductor must be zero and the charges are free to move until this condition is met. The discussion also touches on the importance of using Gauss' law to determine the electric field and enclosed charge.
Physicslearner500039
Homework Statement
In Fig. 23-50, a solid sphere of radius a = 2.00 cm is concentric with a spherical conducting shell of
inner radius b = 2.00a and outer radius c = 2.40a. The sphere has a net uniform charge q1 = +5.00 fC; the shell has a net charge q2 = -q1' What is the magnitude of the electric field at radial distances (a) r = 0, (b) r = aI2.00, (c) r = a, (d) r = 1.50a, (e) r = 2.30a, and (f) r = 3.50a? What is the net charge on the (g) inner and (h) outer surface of the shell?
Relevant Equations
NA

This is my attempt, i am confused at some points
a. r = 0; The Electric field is 0

b. At r = a/2.00; I verified the answer and it is non zero, but my understanding is that the net charge should be on the surface of the conductor. Hence the charge q1=5*10^-15 C, should go to the surface of the solid sphere. Hence the net charge below "a" should be 0 and the E field should be "0". But why is that the charge does not move to the surface?
If the answer is non zero, then i did the following calculation,
the charge is proportional to the volume hence
q' = q *r^3/R^3;

E= q'/(4πεο*r^2); => E = q*r/(4πεο*R^3); => E = (5*10^-15*9*10^9*a)/(2*a^3)
E= 45*10^-6/(2*4*10^-4) = 5.625*10^-2;

c. and d was easy.
E = q/(4πεο*r^2); r = a for c, r =1.5a for d;

e. r = 2.30a this is the area between the conducting shell, the charge on the inner shell is q2 = -q1. At this r, the net charge is 0 q2+q1=0; E=0;

f. r = 3.50a ; E=0 since the net charge is 0.
g. inner q2 = -q1

h. outer 0.

But my additional thinking let us say that q2 = -2q1;
then on the inner shell the charge will be q2 = -q1 since it has to compensate the solid sphere charge. The remaining charge of -q1 will be on the outer shell. Am I correct? Please advise.

Physicslearner500039 said:
Hence the charge q1=5*10^-15 C, should go to the surface of the solid sphere. Hence the net charge below "a" should be 0 and the E field should be "0". But why is that the charge does not move to the surface?

The interior sphere is not conductive. They tell you the charge is uniform through the sphere. Also, you will note that in the case of the shell they make a point of telling you explicitly the shell is conductive.

Physicslearner500039
Physicslearner500039 said:
e. r = 2.30a this is the area between the conducting shell, the charge on the inner shell is q2 = -q1. At this r, the net charge is 0 q2+q1=0; E=0;

Also, that is the interior of a conductor. If the E is not 0. The charges are free to move in the field until it becomes 0. The necessity of the field in the interior of a conductor being 0 is how you arrive at the surface charges, so using them to show the field is zero is a bit of circular logic. I think the statement that the field inside a conductor must be zero is the more fundamental point.

Physicslearner500039
Physicslearner500039 said:
But my additional thinking let us say that q2 = -2q1;
then on the inner shell the charge will be q2 = -q1 since it has to compensate the solid sphere charge. The remaining charge of -q1 will be on the outer shell. Am I correct? Please advise.

Really, you shouldn’t have to guess at any of this. Between Gauss’ law and the fact that the field inside a conductor is zero you should be very confident of your answer. If you put a Gaussian surface in between the surfaces of the shell the field must be zero so the enclosed charge must be zero, right?

## 1. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface. It is named after the German mathematician and physicist Carl Friedrich Gauss.

## 2. How does Gauss' Law apply to a conducting rod?

Gauss' Law can be used to determine the electric field around a conducting rod by considering the charge enclosed within a cylindrical surface surrounding the rod. The electric field will be perpendicular to the surface of the rod and its magnitude will depend on the charge density and the radius of the rod.

## 3. Can Gauss' Law be used to find the electric field inside a conducting rod?

No, Gauss' Law only applies to the electric field outside of a conductor. Inside a conductor, the electric field is zero due to the redistribution of charges that occurs in order to maintain electrostatic equilibrium.

## 4. How does the shape of the conducting rod affect the electric field?

The shape of the conducting rod will affect the distribution of charge and therefore the electric field around it. For example, a rod with a larger radius will have a stronger electric field compared to a rod with a smaller radius, assuming the same charge density.

## 5. Can Gauss' Law be used to find the electric field at a point on the surface of a conducting rod?

Yes, Gauss' Law can be used to find the electric field at any point outside of a conducting rod, including points on its surface. This is because the electric field is perpendicular to the surface of the rod and its magnitude can be determined by considering the charge enclosed within a cylindrical surface surrounding the point of interest.

• Introductory Physics Homework Help
Replies
1
Views
893
• Introductory Physics Homework Help
Replies
23
Views
705
• Introductory Physics Homework Help
Replies
6
Views
219
• Introductory Physics Homework Help
Replies
10
Views
870
• Introductory Physics Homework Help
Replies
5
Views
338
• Introductory Physics Homework Help
Replies
26
Views
750
• Introductory Physics Homework Help
Replies
4
Views
589
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
80
• Introductory Physics Homework Help
Replies
1
Views
1K