Calculating Surface Charge Density and Distance in a Conducting Sphere Problem

[langenase]

Homework Statement

A 32 cm diameter conducting sphere is charged to 500 V relative to V=0 and r = infinity.
(a) What is the surface charge density sigma?
(b) At what distance will the potential due to the sphere be only 10 V?

Homework Equations

The Attempt at a Solution

I know that to find the surface charge density I have to find charge/area. (Sigma = q/A) Finding the area just means finding the surface area of the sphere right? 4(pi)((0.16m)^2)
Is the charge 500 V? It seems like charge should be expressed in coulombs. I'm not really sure how to get the charge of the sphere.


Then to find the distance at which the potential is only 10 V
Do I use V= (1/4(pi)(eo)* sigma/r? where r is that distance and V = 10V?

 

[learningphysics]

Hint: The field outside the sphere is the same as the field due to a point charge where the center of the sphere is... where the point charge is the total charge on the sphere...

Using this, you can find the total charge on the sphere... then the charge density.

 

[Dick]

To get the charge, remember Gauss' law?? The rest of what you're doing seems about right. Remember you are given the sphere diameter in cm not m.

 

[langenase]

The field outside the sphere is given by the equation E=k(Q/r^2)
I solved for Q and got units in C*m. Was it wrong for me to set E=500V?

 

[learningphysics]

The field outside the sphere is given by the equation E=k(Q/r^2)

Cool. That's right.

I solved for Q and got units in C*m. Was it wrong for me to set E=500V?

Yeah, that's wrong... what's the voltage due to a point charge... taking the voltage at r = infinity to be 0.

 

[Dick]

Once you've straightened all of this out, you may want to think about the Gauss law approach. Charge contained in a sphere at the surface is (500V)*A*e0 where A is the area. To convert that to surface charge just divide out the A. Using the point charge will (of course) give you the same thing. It's just a little bit longer way around.

 

[learningphysics]

Once you've straightened all of this out, you may want to think about the Gauss law approach. Charge contained in a sphere at the surface is (500V)*A*e0 where A is the area. To convert that to surface charge just divide out the A. Using the point charge will (of course) give you the same thing. It's just a little bit longer way around.

But the charge is E*A*e0 according to Gauss law... I'm probably missing something...

 

[Dick]

But the charge is E*A*e0 according to Gauss law... I'm probably missing something...

Sorry. I meant surface charge density, which is what the OP is asked for.

 

[learningphysics]

Sorry. I meant surface charge density, which is what the OP is asked for.

But the 500V is voltage not electric field...

 

[Dick]

But the 500V is voltage not electric field...

Right, J/C not N/C. Please ignore me, I'm apparently having a fuzzy day. Thanks.

 

[learningphysics]

Right, J/C not N/C. Please ignore me, I'm apparently having a fuzzy day. Thanks.

No prob. Your fuzzy days are quite rare from what I've seen of your posts.

 

[langenase]

I'm a little confused about what to set E equal to. The voltage due to a point charge is when r is infinity is 0. But how does that relate to the electric field?

 

[learningphysics]

I'm a little confused about what to set E equal to. The voltage due to a point charge is when r is infinity is 0. But how does that relate to the electric field?

Voltage due to a point charge = kQ/r.

You have voltage... you have r, you can solve for Q.