Electrostatics, force applied on a charge

[fluidistic]

Homework Statement

First of all, in order to avoid any confusion, let me precise that I use Gaussian units. I.e. k=1 rather than $\frac{1}{4\pi \varepsilon _0}$.
An infinite conductor plane is kept at a potential worth 0. A point-like charge can be found at a distance d from the plane. Using the method of images, find:
1)The induced surface charge density over the plane.
2)The force that the plane exerts on the point-like charge, integrating the force that each surface elements of the plane exert over the point-like charge.
3)Show that the exerted force by the plane over the point-like charge is equal to the force between the charge and the image.
4)The required work to move the charge q from its position up to infinity.
5)The potential energy between the charge q and the image (compare with the previous result)

Homework Equations

Lots.

The Attempt at a Solution

I've done part 1 and cannot get the right answer for part 2.
For part 1), I reached $\sigma (x,y)=-\frac{qd}{2\pi} (x^2+y^2+d^2)^{-3/2}$. My professor reaches the same.
Now for part 2) I work with cylindrical coordinates (for the shape of the function sigma). This makes $\sigma (r) =-\frac{qd}{2\pi} (r^2+d^2)^{-3/2}$.
In my case an element of area of the plane will be an annulus of radii r and r+dr. Thus in this case their charge is worth $2\pi r \sigma (r)$.
Now I use Coulomb's law to get the force between the plane and the point-like charge and I write the differencial of r as (dr):
$$dF=\frac{q \cdot 2\pi r \sigma (r)}{\sqrt {d^2+r^2}}(dr)=-\frac{q^2dr}{r^2+d^2}(dr)$$.
Now I have to integrate this from r=0 to $r=\infty$. But this integral diverges between the integrand goes as ~1/r.
The answer to part 2) should be $F=-\frac{q^2}{4d^2}$. I've been stuck for days now on this problem, I asked 2 friends that already passed the course and they couldn't spot any mistake in what I've done so far. One had said I should always use vectors and not modulos of the "dF" but I think this shouldn't change my result. The area elements I consider always sum up a contribution, never substract anything.
Any idea would be awesome.

 

[Steely Dan]

One had said I should always use vectors and not modulos of the "dF" but I think this shouldn't change my result.

Of course it will. The assumption you make in writing down your $dF$ is that all of the magnitude of the force from each part of the ring algebraically sums up with every other part, but this cannot be true, because the horizontal part of the force cancels out.

 

[fluidistic]

Of course it will. The assumption you make in writing down your $dF$ is that all of the magnitude of the force from each part of the ring algebraically sums up with every other part, but this cannot be true, because the horizontal part of the force cancels out.

First of all thanks a bunch!
Hmm I understand what you mean but I don't really see it mathematically.
So basically I should take $dA=2\pi r dr d \theta$? So that $dQ=-\frac{qdr(dr)(d\theta )}{(r^2+d^2)^{3/2}}$.
But in this case when I integrate, the only difference I get from my integral in my last post is an extra factor of $2 \pi$. I do not see where my mistake lies.

 

[Steely Dan]

No, I don't mean you need to change your integral. I mean that you need to separate out the two components of the force. If you take a given area element with width $r\ d\theta$, it will have a field component at point $z = d$ that points horizontally and a field component that points vertically. Specifically,

$$F_h = |F| \text{cos}(\phi)$$
$$F_v = |F| \text{sin}(\phi)$$

I am defining $\phi$ here as the angle between the force vector and the $z = 0$ plane. Clearly this angle depends on the value of $r$, so it affects your integration. Now, you know from symmetry that the horizontal components must all cancel, so you don't need to evaluate that integral. But when you are evaluating the vertical component of the force, you do need to take into account that not all of the total Coulomb force is going to be directed vertically.

 

[fluidistic]

No, I don't mean you need to change your integral. I mean that you need to separate out the two components of the force. If you take a given area element with width $r\ d\theta$, it will have a field component at point $z = d$ that points horizontally and a field component that points vertically. Specifically,

$$F_h = |F| \text{cos}(\phi)$$
$$F_v = |F| \text{sin}(\phi)$$

I am defining $\phi$ here as the angle between the force vector and the $z = 0$ plane. Clearly this angle depends on the value of $r$, so it affects your integration. Now, you know from symmetry that the horizontal components must all cancel, so you don't need to evaluate that integral. But when you are evaluating the vertical component of the force, you do need to take into account that not all of the total Coulomb force is going to be directed vertically.

Thanks again. I get $F_v=|F| \cos \phi$ using your definition of the angle phi.
As for the dependence of this angle with r, I get $\phi = \arctan \left ( \frac{r}{d} \right )$. Am I doing things wrong? :/

 

[Steely Dan]

The vertical component should definitely have $\text{sin}(\phi)$ with my definition, since at $\phi = 90^\circ$ the force should only have a vertical component (as the force vector is perpendicular to the plane). But if you want to define it the other way, that's totally fine.

Anyway, the goal here is not to actually evaluate $\phi$. It is possible to write down the value of $\text{sin}(\phi)$ only in terms of $r$ and $d$.

 

[fluidistic]

The vertical component should definitely have $\text{sin}(\phi)$ with my definition, since at $\phi = 90^\circ$ the force should only have a vertical component (as the force vector is perpendicular to the plane). But if you want to define it the other way, that's totally fine.

Anyway, the goal here is not to actually evaluate $\phi$. It is possible to write down the value of $\text{sin}(\phi)$ only in terms of $r$ and $d$.

Ah ok! Well I'll stick with "my" phi angle then :) $F_v= F\cos \phi =F\frac{d}{\sqrt {r^2+d^2}}$.

 

[Steely Dan]

Yes, you've got it. Now try doing the integral and see what you get :)

 

[fluidistic]

Yes, you've got it. Now try doing the integral and see what you get :)

Hmm I don't think I've got it yet .
I had $dF=\frac{q \cdot 2\pi r \sigma (r)}{\sqrt {d^2+r^2}}(dr)$. Now I must multiply this by $\frac{d}{\sqrt{r^2+d^2}}$ to calculate only the vertical component of the force.
When I do this I get $dF_{\text{vertical}}=-\frac{q^2d^2r}{(r^2+d^2)^{5/2}} (dr)$. Integrating this from r=0 to infinity gives me $F_{\text{vertical}}=-\frac{q^2}{3d}$.

Edit: I think I've made a mistake in my first post when I wrote $dF=\frac{q \cdot 2\pi r \sigma (r)}{\sqrt {d^2+r^2}}(dr)=-\frac{q^2dr}{r^2+d^2}(dr)$. The term on the right should be $-\frac{q^2dr}{(r^2+d^2)^2} (dr)$. Integrating from 0 to infinity, the integral does not diverge! I get $F=-\frac{q^2}{2d}$. Which isn't the good result but does not diverge anymore.
I'm turning crazy I think.

 

[Steely Dan]

The force on the test charge from the surface is, by Coulomb's law,

$$dF = \frac{q\ dq}{r^2+d^2} \text{cos}(\phi)dr = - \frac{q^2\ d\ r}{(r^2+d^2)^{5/2}}\text{cos}(\phi)dr = \frac{q^2\ d^2\ r}{(r^2+d^2)^3}dr$$

Make sure you agree with this.

 

[fluidistic]

The force on the test charge from the surface is, by Coulomb's law,

$$dF = \frac{q\ dq}{r^2+d^2} \text{cos}(\phi)dr = - \frac{q^2\ d\ r}{(r^2+d^2)^{5/2}}\text{cos}(\phi)dr = \frac{q^2\ d^2\ r}{(r^2+d^2)^3}dr$$

Make sure you agree with this.

Expect for a sign in the last equality, I agree on this. I finally reach the desired result thanks to you. I had forgotten to square the denominator... I must be more careful. I'll continue this problem later (overloaded with other courses :/). Thanks a lot so far for all your help.