# Calculate the induced charge on a conducting surface

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1. Nov 25, 2016

### doktorwho

1. The problem statement, all variables and given/known data
A very thin plane of length $2a$ is placed in the air at height $a$ above the conducting surface. The plane is charged on its surface and the expression for it's surface charge density ($ρ_s$) is given by $ρ_s=ρ_{s0}*\frac{x}{a}$ and $ρ_{s0}$ is some constant.
a) Determine the surface charge density function inducted on the conducting surface.
The image that goes with this is uploaded.
2. Relevant equations
$$ρ_s=ρ_{s0}*\frac{x}{a}$$
3. The attempt at a solution
I need help in starting to solve the problem. I have some theories as to how i would solve it and i would like a feedback on if it's ok to do so.
I was gonna use a theory where we remove the conducting surface and add a mirror image of the object with an opposite charge. So at $2a$ distance from the first plane there is a second plane with a charge density $$ρ_{s2}=-ρ_{s0}*\frac{x}{a}$$. Now all i have to do is calculate the electric field right? I am confused as how im gonna do this.
I know that the electric field of a plane is $$E_p=\frac{ρ_s}{2ε_0}$$ but how will i find the inducted charge and express it in a surface charge density function?

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2. Nov 25, 2016

### BvU

That's right. As if the conducting plane at z=0 were absent. From symmetry (that's why you use the mirror image method in the first place) you know you only need one component. You need a relevant equation for $\vec E$, something like $|E| = {kq\over r^2 }$ and an integration.

Once you have E, then you use $E_p=\frac{ρ_s}{2ε_0}$ to get the surface charge on the conducting plane at z=0.

If the given charge distribution on the thin plane is causing you a problem, think of a negative charge on one end and a positive on the other: with the mirror you get a quadrupole field. So for this exercise you expect something in that ball park as well.

3. Nov 25, 2016

### doktorwho

Thanks, as for the relevant equation for $E$ in order to get what i said was the $E$ for the plane a used the Gausses's law :-). I understood this problem and m about to post another thats been bothering me :D. Thanks for the answer, your the kind that makes this site awesome (gives a big thumbs up)

4. Nov 25, 2016

### BvU

Gauss is valid for a pillbox close to the plane in case of non-uniform surface charge.

5. Nov 25, 2016

### doktorwho

Hmm..thinking about it now i can use the electric field of a line to replace this plane. What i mean is:
$$E=\frac{ρ_l}{2πε_0r}$$ is the elekctric field of a line and $ρ_l$ is the line charge function. Here it's just $ρ_l=ρ_sdx$. I guess this should work?

6. Nov 25, 2016

### BvU

It didn't appear in the problem statement, but I suppose the charged plane extends infinitely in the y+ and the y- direction, so: yes.

7. Nov 30, 2016

### doktorwho

Sorry for the delay, i was preoccupied but i did make some progress with this and want to verify if this is correct.
So i used a mirror theorem in which i placed at distance $2a$ a plate that is exactly the same as the first one but with a negative charge on it. The next is my work:
$dE_1=\frac{ρ_l}{2\piε_0r}$
$dE_{total}=2dE_1cosθ$
I have choosen some point $x_0$ in where i calculate the field
$cosθ=\frac{a}{r}=\frac{a}{\sqrt{(x_0-x)^2+a^2}}$
when i apply the given surface charge density:
$E_{total}=\frac{ρ_{s0}}{πε_0}\int_{-a}^{a}\frac{xdx}{(x_0-x)^2+a^2}$
This should be the electric field, right? But when i integrate i will get a $x_0$ term and that appears nowhere in the result that my book gives, which i can post if you want..

8. Nov 30, 2016

### BvU

$x_0$ is the x-coordinate of the point (line) on the conducting plate at z=0 where you want to express the surface charge density. I would expect that charge to depend on such an $x_0$ !?

9. Nov 30, 2016

### doktorwho

View attachment 109693
Here is the solution, can you make something out of it?

Last edited: Nov 30, 2016
10. Nov 30, 2016

### BvU

No, except: His/her $x$ is your $x_0$.