Calculating Capacitance Between Parallel Wires & Plane

• Miike012
In summary, you need to find the capacitance between two infinitely parallel long wires that are parallel to a large flat metal sheet. Both plane and lines have uniform charge. Left line: Negative charge densityRight Line: Positive charge densityDistance between lines: LDistance between lines and plane: DDistane between lines and plane: DNow solve for C: C = ΔV(0,0) +∫Epdy|(0,0)Finding capacitance:C = Q/ΔV
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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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.

Simon Bridge said:
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.

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:
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.

Simon Bridge said:
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.

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:
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.

Simon Bridge said:
Can't tell, did you keep track of the charge density?
You'll probably have to do an integration at some point.

Also see:

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

What is the capacitance between parallel wires?

The capacitance between parallel wires refers to the ability of the wires to store electric charge when a potential difference is applied between them. It is measured in farads (F).

How do you calculate the capacitance between parallel wires?

The capacitance between parallel wires can be calculated using the equation C = (ε0*εr*A)/d, where C is the capacitance, ε0 is the permittivity of free space, εr is the relative permittivity of the material between the wires, A is the area of overlap between the wires, and d is the distance between the wires.

What is the significance of the distance between parallel wires in calculating capacitance?

The distance between parallel wires plays a major role in determining the capacitance between them. As the distance decreases, the capacitance increases, indicating a stronger ability to store electric charge. Conversely, with a larger distance, the capacitance decreases.

What happens to capacitance if the area of overlap between parallel wires is increased?

If the area of overlap between parallel wires is increased, the capacitance also increases. This is because there is a larger surface area for charge to accumulate, resulting in a higher capacitance value.

Is the capacitance between parallel wires affected by the material between them?

Yes, the material between the parallel wires, also known as the dielectric material, can affect the capacitance value. Materials with a higher relative permittivity (εr) will result in a higher capacitance, while materials with a lower εr will have a lower capacitance value.

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