# Capacitance between a wire and a plane

1. Jun 21, 2017

### Kelly Lin

1. The problem statement, all variables and given/known data

2. Relevant equations
I want to know whether my process is correct?
THANKS!!!

3. The attempt at a solution

1. By using Gauss's law:
$$E\cdot 2\pi rz = \frac{\lambda z}{\epsilon_{0}}\Rightarrow E=\frac{\lambda}{2\pi\epsilon_{0} r}$$
In my coordinate system, $$E=\frac{\lambda}{2\pi\epsilon_{0}[y^{2}+(z-d)^{2}]^{-1/2}}$$
Then,
$$V=-\int\mathbf{E}\cdot d\mathbf{l}=-\int_{0}^{z}\frac{\lambda}{2\pi\epsilon_{0}[y^{2}+(z-d)^{2}]^{-1/2}}dz=\frac{\lambda}{2\pi\epsilon_{0}y}\ln{\frac{\left | \sqrt{d^{2}+y^{2}}-d \right |}{\left | \sqrt{(z-d)^{2}+y^{2}}+(z-d) \right |}}$$
Since $$C=\frac{Q}{V}$$, then
$$C(\text{per length})=2\pi\epsilon_{0}y\ln{\frac{\left | \sqrt{(z-d)^{2}+y^{2}}+(z-d) \right |}{\left | \sqrt{d^{2}+y^{2}}-d \right |}}$$
2.
Use $$\sigma=-\epsilon\frac{\partial V}{\partial z}|_{z=0}$$ to get the answer.

2. Jun 21, 2017

### haruspex

No, several errors.
You have written the expression for the field at some arbitrary point, (y, z).
First problem: the expression would be correct without the plate. Perhaps the charge distribution on the plate changes it?
Second problem: you integrate this wrt z,but leaving y as arbitrary. What path is that along?
Note that it would be simpler to run the integral in the other direction, avoiding the d-z.
Third problem: if you fix the second problem, you will get an integral that diverges at the wire end. You need to take into account the radius of the wire.
Fourth problem. You have z as a bound on the integral. What should it be?

3. Jun 21, 2017

### Kelly Lin

Now, I correct my process!
Firstly, the electric field generated by the plane is $$E_{plane}=\frac{\sigma}{2\epsilon_{0}}\hat{\mathbf{z}}$$.
Then, by the definition of potential, I can get the potential $$V_{plane}=-\frac{\sigma}{2\epsilon_{0}}z$$
On the other hand, the electric field generated by the wire is $$E_{wire}=\frac{\lambda}{2\pi\epsilon_{0}r}\hat{\mathbf{r}}$$
Then, the potential is $$V_{wire}=-\frac{\lambda}{2\pi\epsilon_{0}}\ln{\frac{r}{a}}=-\frac{\lambda}{2\pi\epsilon_{0}}\ln{\frac{\sqrt{(z-d)^{2}+y^{2}}}{a}}$$
Thus, the total potential will be $$V_{total}=-\frac{\sigma}{2\epsilon_{0}}z-\frac{\lambda}{2\pi\epsilon_{0}}\ln{\frac{\sqrt{(z-d)^{2}+y^{2}}}{a}}$$

Now, we can use the relation $$\sigma=-\epsilon\frac{\partial V}{\partial z}|_{z=0}$$ to get the surface charge density.
$$\sigma = \frac{\lambda}{2\pi} \frac{a}{\sqrt{(z-d)^{2}+y^{2}}} \frac{1}{2} \frac{2(z-d)}{\sqrt{(z-d)^{2}+y^{2}}} |_{z=0} + \frac{\sigma}{2}\\ \frac{\sigma}{2} = \frac{\lambda}{2\pi} \frac{-ad}{d^{2}+y^{2}}\\ \sigma = \frac{-\lambda a d}{\pi(d^{2}+y^{2})}$$
Then, we can go back and get the total potential in the system
$$V =-\frac{\lambda}{2\pi\epsilon_{0}}\ln{\frac{\sqrt{(z-d)^{2}+y^{2}}}{a}} + \frac{\lambda adz}{2\pi\epsilon_{0}(d^{2}+y^{2})}$$
and the capacitance per length of the wire is
[tex]
C = \frac{\lambda}{V} = \frac{2\pi\epsilon_{0}(d^{2}+y^{2})}{adz} - 2\pi\epsilon_{0}\ln{\frac{a}{\sqrt{(z-d)^{2}+y^{2}}}}
[\tex]

4. Jun 21, 2017

### haruspex

That expression is for a uniformly charged plane. It won't be.
Consider the method of images.

5. Jun 21, 2017

### Kelly Lin

At first glance of the question, I came up with this method.
However, the plane isn't grounded so I am unsure if this method is valid for this problem?

6. Jun 21, 2017

### haruspex

It is an infinite plane. That is effectively grounded.

7. Jun 21, 2017

### Kelly Lin

Oh~ I see~
Thanks a lot!!!!!!!!!