Potential of Concentric Spherical Insulator and Conductor --

[scef333]

Potential of Concentric Spherical Insulator and Conductor...Please Help

Homework Statement

A solid insulating sphere of radius a = 4.9 cm is fixed at the origin of a co-ordinate system as shown. The sphere is uniformly charged with a charge density ρ = -108 μC/m3. Concentric with the sphere is an uncharged spherical conducting shell of inner radius b = 12.8 cm, and outer radius c = 14.8 cm.

1) What is Ex(P), the x-component of the electric field at point P, located a distance d = 34 cm from the origin along the x-axis as shown?

What is V(b), the electric potential at the inner surface of the conducting shell? Define the potential to be zero at infinity.

What is V(a), the electric potential at the outer surface of the insulating sphere? Define the potential to be zero at infinity.

Homework Equations

U12=k(q1q2/r12) k=9E9

The Attempt at a Solution

I was able to find the answer to the first question using the equation i gave but when i tried that for 2 and 3 it didnt work and gave the message: "It looks like you have calculated the potential at the inner radius of the shell to be equal to the potential at r = c produced by the insulating sphere by itself. The conducting shell plays a role here. Go back to the definition of the potential to determine the answer."

I don't really understand how the outside uncharged spherical conducting shell can have any effect when it is uncharged. I must be missing something here. Please give input. A diagram is attached.

 

[chw42]

I'm also interested in this question. I got the answer for the potential at v(b), but not at v(c), the outside of the insulating sphere. Thing is, I don't know how the outside shell can have any effect since it has zero charge. I know that the inside of the shell will be negatively charged while the outside is positive though.

 

[chw42]

I figured it out.

You have to account for it, not totally sure why, but to find out the potential of the ring it's KQ(1/b - 1/c), then you add that to the potential at the location you want.

 

[Enterprise]

It looks like you have calculated the potential at the outer radius of the insulating sphere to be equal to the potential at r = a produced by the insulating sphere by itself. The conducting shell plays a role here. Go back to the definition of the potential to determine the answer.

if i use a=r, this is what i get.
which point to take as r?

 

[Sneakatone]

how did you solve #1? I am guessing you replaced q1 and q2 and used density to solve for q. if you did did you use 4/3*pi*r^3(-108) to get it? (so should r be the inner or outer radius?)

 

[Sneakatone]

never mind I used the radius a to solve for q

 

[cello_mike]

I figured it out.

You have to account for it, not totally sure why, but to find out the potential of the ring it's KQ(1/b - 1/c), then you add that to the potential at the location you want.

The reason this works is because there is no electric field in the ring. chw42 must have botched a sign, because you want to subtract the contribution of the uncharged ring from the total electric potential at point B. $$\frac{KQ}{b} - \frac{KQ}{b-c}$$

It would probably help to start from the integral in order to keep track of signs and see where they cancel out.

 

[ehild]

What is the electric field between the sphere and the shell at distance a<r<b? If there is some electric field between a and b, there must be also some potential difference. What is it?

Can you give the potential at the outer surface of the shell? It is a metal shell, what is the potential at any point of it? So what is the potential on the inner surface?

ehild

 

[Future Aero Engineer]

To find the V(b) the potential on the inner surface of the shell you will need two integrals. The first integral will be from infinity to the outer surface, and the second will be from the outer surface to the inter surface. The electric field for the first integral is the total field from both charges (image drawing a gaussian surface around the whole system of spheres) and the electric field for the second will be zero because the field is zero inside conductors. This leaves you with only the first integral that integrated will give you that the potential is (KQ)/r , where r is the radius from the center of the sphere to the outer surface of the conducting shell. I just did this problem myself and hope this helps anyone else! :)