Electric Force on a charged conductive bodyby juvan Tags: body, charged, conductive, electric, force 

#1
Dec712, 08:51 AM

P: 12

Hi,
how to calculate the force on a charged conductive body (ball), when that body is between two parallel plates (+V, V) (see image), ooh and I am calculating numerically. I would first calculate the surface charges on the conductive surfaces (2 disks + ball), then calculate the potential field from the two disks ONLY (with those surface charge configurations, where the charged ball also had a say), and from this field, go on to calculate E=∇V. I think the ball can't exert the force on itself (kinda like trying to pick yourself up while standing in a basket and pulling up on the handles), that is why I think you would only need to calculate the field from the disks, and then with this field E, calculate the force on the body F=∫dQ*E. Would you agree, or am I missing something. I get strange values of equipotential lines in the contour plot (see image). PS: One more question, if you charge a ball to a potential +V (touch on of the disks) and then move the ball away from that disk (no longer touching) towards the second disk, would the potential of the ball change on its way?? Thanks all. 



#2
Dec712, 09:04 PM

P: 1,027

Hello juvan,
your ideas are correct, only the electric field due to plates is neccessary. However, in order to calculate the charges, you will most probably need to find the potential first. This you can find from Poisson's equation, with boundary conditions on the plates and the object between them. The equipotential planes should reflect the presence of the conducting body between the plates  they should be spheres surrounding the conducting ball. 



#3
Dec812, 04:06 AM

P: 12

Thank you Jano, for your reply. Now let me further elaborate what I did. I started all the work from the solution of Poisson's equation for the electric potential [itex]V(T)=\frac{1}{4\pi\epsilon_0}\int\limits_A \frac{\sigma(T')da'}{R}[/itex] then I went on to copy some of the "magic" from the book, to set up the equations, and then with that set up I used the MoM to solve the integral equation for the surface charge density on the conductors [itex]\sigma[/itex], where I used  to generate all the equations for all the unknowns  the known potentials on the conducting surfaces (2 disks+ball). So now with the surface charge distributions known, I could easily go back to calculate the electric potential, and I got (see attached image). BUT I think that something is not completely correct so far, I have a hunch that I can't just say that the potential of the ball in the field is equal to some constant value (the value she got when she touched one of the disks).
Just to once more for clarity tell what my job is, I have to calculate the electric force exerted on the ball in between two charged parallel disks. Thank you still ^^ 



#4
Dec812, 07:10 PM

P: 1,027

Electric Force on a charged conductive body
Can you post the whole text of your task?
If the ball touched one of the planes (left I guess, since it has the charge of same sign?), it has the same potential as the plane. Some charge will be transferred to/from the ball. Removing ball away will leave the charge on the ball the same, and the potential on the ball is everywhere the same. However, it won't be equal to plate potential anymore. How did you find out the charge distribution first? I think first one has to solve Poisson's equation numerically to find the potential everywhere. Then electric intensity, then charge. 



#5
Dec1012, 04:41 AM

P: 12

Well there isn't really any text to the task, just the task itself, which is to calculate what would happen to a charged ball in between two conducting disks at some potential. That's it. So i, from what I know (the potential), said let the ball be touching one of the disks so I know all of the conductive surfaces potentials, and I can go on to calculate all the surface charge distributions. That is what I did, but the problem I run upon is the one you also said that




#6
Dec1012, 05:16 AM

P: 1,027

OK. You are right, the charge densities will move when the ball is displaced.
Let's denote the two situations in this way: 1: ball touches the left plate 2: ball does not touch any plate I am not sure how you found out the charge with the above equation in 1. We know only the potentials on the surfaces, not in the space between them, and the former are not sufficient. Surface charge density can be found from the Gauss law as dot product of the electric field just above the surface and the normal vector to the surface: [tex] \sigma = \mathbf E \cdot \mathbf n. [/tex] In order to do that, we have to know [itex]\mathbf E[/itex]. This can be calculated from the potential: [tex] \mathbf E = \nabla V. [/tex] So in fact we have to find potential first. There is Poisson's equation [tex] \Delta V = \frac{\rho}{\epsilon_0} [/tex] which has to be solved with these conditions:  V(left plate) = V_1  V(ball) = V_1  V(right plate) = V_2  V(infinity) = 0 When V is found as a function of position, one can get back and calculate the electric field and only then the charge [itex]Q_B[/itex] on the ball. In situation 2, one has to do the calculation of new potential V', but now with the ball in different place and with conditions  V(left plate) = V_1  V(right plate) = V_2  V(infinity) = 0 and [tex] \oint_S \nabla V' \cdot d\mathbf S = Q_B, [/tex] where the integration is over surface enclosing the ball. 



#7
Dec1012, 06:11 AM

P: 12

Hmm.. Ok. Firstly I'll just tell you how I calculated σ only from the potentials at the cond. surf. If you look at the equation I wrote for V(T), I will rewrite it now for the way I did it numerically.
[itex]V_j = V0 = \frac{1}{4\pi\epsilon_0}\sum_{i=1}^{N} \frac{\sigma_i' dS_i'}{R_{ij}} ; j=1,2,...,N[/itex] So this is one equation with N unknowns (σi...σN), and since you know the potential at all the N points on the surfaces (since you're calculating for σ only on the surfaces), you have therefore N equations and then I just linear solve the Matrix equation A*σ=b for the σ. This is it. And this, I think, is the basic Method of Moments numerical method. So now I'm still stuck with the problem we are discussing, and your previous reply, despite it being well written, doesn't help me get unstuck xD Thanks for all the time you took to help me so far. 



#8
Dec1012, 08:57 AM

P: 1,027

Aah I see now, I should have read your post more carefully. This method of determining potential is new to me, and I like it, it's much easier than to solve some partial dif. equation.
I think you can go on in this way: We want to find the force on the ball. This is given by [tex] \oint_{surface~of~ball} \sigma \mathbf E_{ext} dS [/tex] where [itex]\mathbf E_{ext}[/itex] is the field due to plates. It is not the total field (this is even discontinuous on the surface). In order to find it, we have to find the charge density on the plates and then use Coulomb's law. We need the charge density on the ball as well. You can find [itex]\sigma[/itex]'s in the same way you explained above. Set up linear equations tying N+M potentials at different points of the plates to N surface charges on the plates and to M surface charges on the ball. This time, however, for potentials use only points which are on the plates, not those on the ball, as the potential there is not known. Such system of equations has infinity of solutions. This is because the so far the net charge on the ball could have any value. But we know that the surface charges on the ball sum up to [itex]Q_B[/itex]; this gives us one more equation [tex] \sum_{i\in~ball} \sigma_i \Delta S_i = Q_B, [/tex] which should make the above system to have a unique solution. 



#9
Dec1112, 06:05 AM

P: 12

But I think I somehow misunderstood you. But I read the post like 10 times, and I can't understand your logic. Thanks. PS: a side question. If you take a battery and connect it to two conducting surfaces, what does the battery do? (I think that what it does, it redistributes the charge (doing some work in the process Battery voltage) from the positive (takes electrons) and carries it to the negative side (dumps electrons there). So by this logic the charge on the + side should be +Q and the charge on the negative should be Q " [itex]+QQ=0[/itex] ". So now to transfer the problem to my domain, If at the beginning when the ball is touching the left plate and we connect the battery this equation should hold [itex]Q_{left plate} + Q_{ball} = Q_{right plate}[/itex]). Is this the case or is it not?? (if I am right then my numerical result is wrong at the very beginning EDIT: (actually I just reran it, and it's off by about 0.7% which is actually pretty good i guess. Guess I spoke too soon) 



#10
Dec1112, 07:32 AM

P: 1,027

I meant this: introduce N+M points [itex]\mathbf x_j[/itex] somewhere on the plates to define potential [itex]V(\mathbf x_j)[/itex] (let's use index [itex]j[/itex] for these potential points only). Then introduce N+M points [itex]\mathbf r_i[/itex] (radius vectors) to define positions of the charges, both on the plates (N) and on the ball (M). The sets [itex]\{\mathbf x_j\}[/itex], [itex]\{\mathbf r_i\}[/itex] should be disjunct(no common point), otherwise some distance [itex]R_{ji} = \mathbf x_j  \mathbf r_i[/itex] would be zero, which would be a problem.
Then we have this system of equations: [tex] V(\mathbf x_j) = \sum_{i = 1}^{N+M} \frac{1}{4\pi\epsilon_0} \frac{\sigma(\mathbf r_i)}{R_{ji}}, [/tex] which consists of N+M equations for N+M unknown sigmas. Your thinking about the transfer of charge seems right, the sum of charges at left plate, ball and right plate should be zero, otherwise some charge would get lost in the battery, which does not seem possible. 



#11
Dec1112, 08:46 AM

P: 12

Ok, this is exactly how I calculate all the [itex]\sigma(r_i)[/itex] when I know all the potentials, BUT I cannot see how you meant to calculate the potential of the ball once it is no longer touching the plane, with this equation. You were leaning on the fact that we now know [itex]Q_{ball}[/itex], but how do we use this fact?
EDIT: the assistant professor replyed like this to my question about the potential when the ball is no longer touching the plate.: Ans: When the ball is no longer touching the plate, the potential is in fact not know, but what we do know is its charge. 



#12
Dec1112, 09:07 AM

P: 1,027

I did not do the calculation myself, but it seems natural that it has infinity of solutions. However, we know the charge on the ball and the fact that the total charge is zero. Mathematically, we include equations [tex] \sum_{i \in ~ball}\sigma_i \Delta S_i = Q_B [/tex] [tex] \sum_{i =1}^{N+M} \sigma_i \Delta S_i = 0. [/tex] to the system. Then the system should have unique solution [itex]\{\sigma_i\}[/itex]. The potential on the ball will remain unknown, but we do not need it for the calculation of force. 



#13
Dec1212, 02:50 AM

P: 12

Back again. Ok this time I'll take an overall view of the problem to try to explain as best as I can where I get stuck. Let's word the problem again. Bare with me x)
There are two identical charged disks distance [itex]d[/itex] apart. Calculate the force exerted on a conductive ball, when it is put between these two disks, for simplicity lets asume we start by placing it so it is touching the left disk. So we have come some way, but we (I) are stuck at the problem, what to do when we move the now charged ball, away from the disk. Let me guide you through my tought process so you can hopefuly see where I get stuck: I CAN calculate all the σ (with the equation we set up in previous posts) on all the surfaces (ball(M) + disks (2N)) WHEN I know ALL (2N+M) the potentials. So that's why I start with the ball touching. BUT now as I move the ball away I am left with only 2N (N on one disk) known potentials, and 2N+M (M on ball) unknowns (all the σ's), without the know potentials on every slice on the ball (all the σ that go from 2N+1 to M) I don't know how to get all the σ. And I think I have to have all the σ to be able to calculate the net force on the ball. So the big question is where do we get the M equations needed to uniquely solve for all [itex]\sigma[/itex]. Now.. the one information I sense has value is the TOTAL [itex]Q_{ball}[/itex] charge on the ball, which we took with us, when leaving the plate. But this is still not a lot of information to go on, at least I can't see it. I hope you understand where I'm still stuck. If you can read Mathematica code, I would be happy to provide it... perhaps, if it helps. The upper case of the words may be overkill, but I just like to emphasize it for clarity. I attached a picture, just for some added clarity. 



#14
Dec1212, 08:31 AM

P: 1,027

I think I understand you. It seems that you've got stuck because you want to use the same space points (on the plates) to introduce the charge densities [itex]\sigma_i[/itex] and the potentials [itex]V_j[/itex].
But this is not necessary (and as you see, won't give you enough equations). Instead, you can introduce two completely distinct sets of points: N + M potential points [itex]\mathbf x_j[/itex] on the left surface; these are in potential V_1; N + M potential points [itex]\mathbf x_j[/itex] on the right surface; these are in potential V_2; (the potential is due to all charges) 2N + 2M charge points [itex]\mathbf r_i[/itex] (2N on the plates and 2M on the ball); these are unknown. So together, there are 2(2N+2M) distinct space points. There are 2N+2M known potentials and 2N+2M unknown charges, so there is hope to solve for the unknowns. Is it more clear now? 



#15
Dec1312, 05:23 AM

P: 12

Not really. Could you just write out the equations, that should probably tell me more than words.




#16
Dec1312, 03:25 PM

P: 1,027

Use totally 4(N+M) different points:
N + M points [itex]\mathbf x_j[/itex] on the left plate, all at potential [itex]V(\mathbf x_j) = V_1[/itex] N + M points [itex]\mathbf x_j[/itex] on the right plate, all at potential [itex]V(\mathbf x_j) = V_2[/itex] N points [itex]\mathbf r_i[/itex] on the left plate, N points [itex]\mathbf r_i[/itex] on the right plate, 2N charges [itex]\sigma_i = \sigma_i(\mathbf r_i)[/itex] unknown 2M points [itex]\mathbf r_i[/itex] on the ball, 2M charges [itex]\sigma_i = \sigma_i(\mathbf r_i)[/itex] unknown. So, there are 2(N+M) unknowns ([itex]\sigma_i[/itex]), and 2(N+M) known potentials [itex]V(\mathbf x_j)[/itex]. The system of equations is [tex] V(\mathbf x_j) = \sum_{i = 1}^{2N+2M} \frac{1}{4\pi\epsilon_0} \frac{\sigma(\mathbf r_i) \Delta S_i}{R_{ji}},~~j = 1,...,2(N+M) [/tex] which should have infinity of solutions. Including additional constraint equations [tex] \sum_{i\in ball} \sigma_i \Delta S_i = Q_{ball} [/tex] [tex] \sum_{i\in (ball+ plates)} \sigma_i \Delta S_i = 0 [/tex] should make the system into one with unique solution. 



#17
Dec1612, 05:05 AM

P: 12

Hi again, this time I think I've got it. So to end this thread properly I'll just quickly tell what I did.
I think that I basically did what you said, it's just that I didn't exactly understand it from the equations and things you said. This is the main equation I was starting from [itex]k\sum_{i=1}^{M}\frac{\sigma_i \delta a_i}{R_{ij}}=V_{body}V_0(T)\hspace{0.3cm};j=1,2,...,M[/itex] [itex]k\sum_{i=1}^{M}\frac{\sigma_i \delta a_i}{R_{ij}}=\sum_{i=1}^{M}\sigma_i \alpha_{ik}[/itex] But the thing to note is that the right hand side has Vbody and V0, Vbody is the potential of the body, and V0 is the potential of the external field, in this case, from the 2 disks. So this comes out to M equations BUT there is M+1 unknowns M [itex]\sigma_i[/itex] and one [itex]V_{body}[/itex]. And because we know the total charge on the ball we can use this as the last equation, [itex]\sum_{i=1}^{M}\sigma_i \delta a_i = Q_{ball}[/itex] giving us the final equation to properly solve for all the unknowns like this [itex]\begin{bmatrix} \alpha_{11} & \alpha_{12} & \cdots & \alpha_{1M} & 1\\ \alpha_{21} & \alpha_{22} & \cdots & \alpha_{2M} & 1\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ \alpha_{M1} & \alpha_{M2} & \cdots & \alpha_{MM} & 1\\ \delta a_1 & \delta a_2 & \cdots & \delta a_M & 0 \end{bmatrix} \begin{bmatrix} \sigma_1\\ \sigma_2\\ \vdots\\ \sigma_M\\ V_{body} \end{bmatrix} = \begin{bmatrix} V_0(T_1)\\ V_0(T_2)\\ \vdots\\ V_0(T_M)\\ Q_{body} \end{bmatrix} [/itex] Which is nicely solvable. So this is it. Now I can finally proceed. I can also see now, that we actually don't or I should say, wouldn't need the potential Vball to calculate the force. Thank you Jano, for all your help. 



#18
Dec1612, 08:01 AM

P: 1,027

If I understand your scheme, V0(T) is the potential due to disks at the points on the surface of the ball,
But how do you know this? It is a function of position, so it varies point to point. I think this is unknown, too. Unless the charge distribution on the disks is rigid? 


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