As far as I know Coulomb's law only works for point charges but what if there are not any point charges? For example, let's imagine there are three rectangles with different sizes. First one is 50 cm, second one is 30 cm and last one ise 10 cm. The distance between first and second triangle is 10 cm and the distance between the second and third rectangle is also 10 cm. If their centers are collinear what is the net force acting on second ractangle? Should I calculate it by think their centers as a point charge? I searched for it but couldn't find a clear answer.

Here is the picture of the problem for those who couldn't understand from my explanation.


  • $\begingroup$ I guess you think of point masses, when mentioning Coulomb's Law. Have a look at en.wikipedia.org/wiki/… $\endgroup$ Aug 12 '14 at 6:12
  • $\begingroup$ @Lord_Gestalter I'm trying to understand it. Could you show me how I can apply it to questions? Let's say first one's area is 500 cm^2, second one's is 300 cm^2 and last one's is 100 cm^2. How can I apply it to this question? $\endgroup$
    – Starior
    Aug 12 '14 at 6:56
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    $\begingroup$ If the distances would be big with respect to the distribution of charges you could use the point charge as approximation. This is not given here. Sorry, I'm in a hurry, otherwise it wouldn't have been a comment but an answer ;-) $\endgroup$ Aug 12 '14 at 8:27

It depends how the charges are distributed in the material, and on the material's conductance. If you have a metal, the charges of the plates would be mobile and result in a hard to compute distribution. I cannot help you with that. There are probably good approximations to tackle those kind of problems but I am no expert.

If the charges are static and equally distributed among the surface, and the material has a relative permittivity ($\varepsilon_r=1$), you can use Coulombs law with respect to infinitesimal parts of the charges and integrate over them.

If you suppose that the rectangles have a width of $10cm$, the force of the top plate on the middle plate could be calculated by

$$ \hat{F}_{12} = \frac{1C^2}{4\pi\varepsilon_0}\int_{-25cm}^{25cm}\frac{dx_1}{50cm} \int_{20cm}^{30cm}\frac{dy_2}{10cm} \int_{-15cm}^{15cm} \frac{dx_2}{30cm} \int_{0cm}^{10cm}\frac{dy_2}{10cm} \frac{1}{(x_2-x_1)^2+(y_2-y_1)^2} \begin{pmatrix}x_2-x_1\\y_2-y_1\end{pmatrix} $$

As you can see, this is already quite complicated with the favourable assumptions we made. This should have an analytical solution but at the moment I am too lazy to do it. Wolfram Alpha could probably do the separate integrals and you would have to piece them together. You could then compare the result with what you would expect from point charges.

Ahh, and don't forget that there is the other plate as well. You would need to repeat the integration with opposite sign to obtain the second force and then take the difference for the total force.

  • $\begingroup$ Thanks for the answer. I just want to know that if I simply calculate the force by thinking the centers of the objects as point charges how accurate would my answer be? $\endgroup$
    – Starior
    Aug 12 '14 at 8:41
  • $\begingroup$ As Lord_Gestalter already mentioned, the objects would have to be much farther apart than their physical extent for that approximation to work reliably. In such a case the answer would be rather poor. $\endgroup$
    – M.Herzkamp
    Aug 12 '14 at 8:51
  • $\begingroup$ And one more thing...Can I use Gauss' law like E=ρ/2ε0 to calculate the electric field and then calculate the force by F=qE? $\endgroup$
    – Starior
    Aug 12 '14 at 10:58
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    $\begingroup$ That is only the monopole approximation of the object's field. You would get the same result as assuming the objects as points, because that is what a monopole is: a point. $\endgroup$
    – M.Herzkamp
    Aug 12 '14 at 11:13
  • $\begingroup$ Then is this wrong? I'm so sorry for my ignorance. I'm quite new to this subject. $\endgroup$
    – Starior
    Aug 12 '14 at 11:34

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