Capacitor Physics: Calculating Potential at Point P with Charged Cylinders

[minifhncc]

Homework Statement

[PLAIN]http://img841.imageshack.us/img841/5497/phys.jpg

Capacitor has two cylinders (conducting), as shown.

Charged 0V on LHS cylinder and 12V on the other one. d=140mm.

Show the potential at P is Vp = 6-2.34ln((140-x)/x)

Homework Equations

E=integralE.dA

The Attempt at a Solution

I got as far as using Vp = VPQ + VPR but I cannot seem to get the constant term 6 or the 2.34... could someone please show me the correct method for doing this?

Thanks

 

[ApexOfDE]

Firstly, you find the total electric field at P. From this result, you calculate the potential at point P (from a to d-a). When you arrive this point, consider conditions given in problem: VL = 0 (x = a) and VR = 12 (x = d-a). Plug these into the potential you just found to determine some constants.

 

[minifhncc]

Thanks for your reply. But I can't seem to be able to get this, especially the constant value (ie. 6)... Where does that come from?

 

[ApexOfDE]

Sry for this inconvenience. I will try again.
1/ You should write down the form of electric field caused by a cylinder conductor at point x outside the conductor.
2/ Find total field at point P (beware the sign of RHS field).
3/ Find potential (integrate from a to d-a)
4/ After getting step 3, in the form of V, there is a constant q/2πɛz. This constant contribute with another constant (ln(?)) to be 2 constants in your solution. You will determine it by plugging value V = 12 V at point x = d-a.

 

[minifhncc]

Thanks for your reply...

4/ After getting step 3, in the form of V, there is a constant q/2πɛz. This constant contribute with another constant (ln(?)) to be 2 constants in your solution. You will determine it by plugging value V = 12 V at point x = d-a.

This is where I'm getting stuck :( ... So I should get q/2πɛz = two terms?

Thanks alot

 

[ApexOfDE]

After step 3, the potential at point P is:

V = \frac{q}{2 \pi \epsilon_0 z} [\ln \frac{d-x}{x} - \ln \frac{d-a}{a}]

Now plug V = 12 and x = d-a into this formula, and you will have:

12 = -2 \frac{q}{2 \pi \epsilon_0 z} \ln \frac{d-a}{a}

This will yield constant "6" in your solution.