# Finding capacitance

1. Mar 30, 2014

### Miike012

I need guidance calculating the capacitance between two infinitely parallel long wires that are parallel to a large flat metal sheet.

Condition:
V = 0 at the origin.

Both plane and lines have uniform charge.
Left line: Negative charge density
Right Line: Positive charge density

Distance between lines: L
Distane between lines and plane: D

My attempt.
Find Electric field (El) due to both lines at point P
Find Electric field (Ep) due do plane at point P

E at point P = EP = Ep + El

ΔV(x,y) = -∫EPdy + C
Now solve for C: C = ΔV(0,0) +∫EPdy|(0,0)

Finding capacitance:

C = Q/ΔV
ΔV =
... not sure what to do from here...

Is my approach correct so far?

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2. Mar 30, 2014

### Simon Bridge

Start with neutral wires. The two wires start out at a non-zero potential (because of the charged sheet).

Move charges a bit at a time from one wire to the other one.
Initially this takes no work but gradually the work increases.

What is the relationship between the work to move total charge Q and the capacitance.

3. Mar 30, 2014

### Miike012

ΔV(x) = pL/(piε0)ln[(d-a)/a] = Change in potential from the surface of
pL>0 to pL< 0 wire.
pL: Units C/m

i know you asked for the work but is it ok if i just calculate the potential?

Last edited: Mar 30, 2014
4. Mar 30, 2014

### Simon Bridge

The idea is to use some physics to work out the capacitance.
The work moving the charge is the energy stored in the capacitor ... which has a nice relationship to the capacitance and is general.

But the suggestion is mostly just to nudge you into thinking in terms of physics, so do it whichever way you feel most comfortable.

5. Mar 30, 2014

### Miike012

Ok well the work in moving charge q is
q(pL/(piε0)ln[(d-a)/a])

The total work in moving charge q from the right wire to the left then to the plate is...

q{pL/(piε0)ln[(d-a)/a] + psd/(2ε0)} = qΔV

pL = QL/L
ps = Qs/A ; A = L2

Last edited: Mar 30, 2014
6. Mar 31, 2014

### Miike012

Am I doing your approach correctly?
Actually the work in the y direction should be due to boh the line charges and plate... I didn't include the line charges.

7. Mar 31, 2014

### Simon Bridge

8. Mar 31, 2014

### Miike012

Sorry, I left all the integration out..
This is what I did.
General Eq. for Eline = pL/(2piε0r),
1.r is the distance from the line and
2. pL is the charge density of the line

The given the problem.
Eline = ELeft Line + ERight Line
= pL/(2piε0)[1/(d/2 - x) + 1/(d/2+x)] { points in the neg, x direction}

d/2 - x is the distance from an arbitrary point (x,0) to the right line
d/2+x is the distance from an arbitrary point (x,0) to the left line

Now.. V(x) = -∫Edx = pL/(2piε0)ln[(d/2+x)/(d/2-x)] + C
and
V(0) = 0 = C.
Now the change in potential from the right line to the left line is.
ΔV = pL/(2piε0)ln[(d-a)/a]

Last edited: Mar 31, 2014
9. Mar 31, 2014

### Simon Bridge

Without checking your arithmetic - OK.
So where do you go from there?