How Do You Calculate Electric Potential Inside a Nonconducting Sphere?

[stunner5000pt]

I have two questions, please help i really need to get this done properly!

The electric field inside a NONCONDUCTING sphere of radius R containing uniform charge densiy is radially directed and has magnitude of

E = qr / 4 pi epsilon0 R^3

where r is the distance from the centre

and R is the radius

q is the charge on the sphere

a) Find potential V inside the sphere taking V=0 @ r = 0

Since V = Integral E dr
Then V = qr^2 / 8 pi epsilon0 R^3

then if r = 0 then V = 0

am i correct here?

Show that the potential at distance r from the centre where r <R is given by V = q (3R^2 - r^2) / 8pi epsilon R^3

It looks like it has been integrated from r to (root 3) R. But i don't understand why??


Another problem is

A geiger counter has a metal cylinder of 2.10 dimater with a wire stretched along it's axis whose diamtere is 1.34 x 10^-4 cm in dimater. If 855 V is applied between these two what is the electric field at the surface of the wire and the cylinder??

lets say lambda = Q / L

then flux = EA = E 2 pi r L = 4 pi k Qenc = 4 pi k lambda L

so 2 k lambda / r = E

then i integrate because V = integrate E dr

so that V = 2k lambda Log r

But now i m stumped as to how to proceed please help!
thanks a lot

 

[artybear]

account for the integration constant

This is only a response to your first question about the potential withing a non-conducting sphere. You are right that you can choose the potential at the center of the sphere to be zero. You can set it to whatever you want, since you are generally only concerned with the *change* in potential.

But, what about that sqrt(3) in the equation that some people derive?

Consider the other integral you take to determine the potential of a point *outside* the non-conducting sphere. In that case, it is common to choose the potential at r=infinity to be 0. Now you have two equations that describe the potential near the sphere, based on whether or not it is inside, and presumably, the transition of the potential going from inside the sphere to the outside of the spehre should be continuous. In fact, the slope of the potential at r=R using the equation from inside the sphere equals the slope at r=R from outside the sphere. To make the curves of the two potential equations meet up, we add a constant to the equation from inside the sphere. I hope that helps. I don' t have time right now to write out equations in detail.

 

[wais]

For the first question, your approach and answer seem correct. You are using the equation V = ∫Edr to find the potential at a certain distance r from the center of the sphere. Since the electric field inside a nonconducting sphere is given by E = qr/4πε0R^3, when you integrate this with respect to r, you get V = qr^2/8πε0R^3. And since the potential at r = 0 is given as V = 0, your answer is correct.

For the second problem, it seems like you are trying to find the electric field at the surface of the wire and cylinder using the equation E = 2kλ/r. However, this equation is for a point charge, not a charged wire or cylinder. To find the electric field at the surface of a wire or cylinder, you can use the equation E = λ/2πε0r for a charged wire, and E = σ/ε0 for a charged cylinder, where λ is the linear charge density and σ is the surface charge density.

To find the potential at a distance r from the wire or cylinder, you can use the equation V = ∫Edr, as you mentioned. But in this case, you need to use the equation E = kλ/r for a charged wire, and E = kσ/r for a charged cylinder. So your final equation for the potential would be V = kλln(r) for a charged wire, and V = kσln(r) for a charged cylinder. You can then plug in the values given in the problem to find the potential at the surface of the wire and cylinder.

I hope this helps! If you are still stuck, it might be helpful to review the equations for electric field and potential for different charge distributions. And don't forget to double check the units of your answers to make sure they are in volts for potential and newtons per coulomb for electric field.