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Consider two charged balls in space with radii $R_A$ and $R_B$, and total charge $Q_A$ and $Q_B=\rho_B\space V_B$. They are separated by a distance $D = \|\overrightarrow{AB}\|\geq R_A + R_B$.
I was wondering if it was possible to determine the force exerted by one ball on the other, e.g. $\overrightarrow{F}_{A\to B}$.
First I wasn't sure if it is possible to say that $\overrightarrow{dF}_{A\to B}=Q_B\space dV\space\overrightarrow{E}_A$ where $dV=r^2\space \sin{\theta}\space d\phi\space d\theta\space dr$. Then, if this is correct, it leads to $$\overrightarrow{F}_{A\to B}=\frac{\rho_B\space Q_A}{4\space\pi\space\epsilon_0}\iiint_{V_B}\frac{\overrightarrow{AM}}{\|\overrightarrow{AM}\|^3}dV$$ Is it possible to "calculate" this integral in the general case or is it impossible? Thanks in advance

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  • $\begingroup$ you do not need the integral, balls behave like point charges . The force is $F= kQ_A Q_B/D^2$ $\endgroup$
    – user65081
    Commented May 26, 2020 at 19:17
  • $\begingroup$ What is $\overrightarrow{AM}$ supposed to mean? $\endgroup$
    – G. Smith
    Commented May 26, 2020 at 19:35
  • $\begingroup$ @Wolphramjonny It isn’t obvious that balls with spherically-symmetric charge densities behave like point charges. It is just a well-known result of doing this kind of integral... typically for the potential of a shell and then reasoning from there. $\endgroup$
    – G. Smith
    Commented May 26, 2020 at 19:38
  • $\begingroup$ @G.Smith, well, but I assume he knows gauss' theorem $\endgroup$
    – user65081
    Commented May 26, 2020 at 22:16
  • $\begingroup$ @Wolphramjonny If you know Gauss’ Law, I agree that you can use it to easily derive the field of a spherically symmetric ball without integrating. But I don’t think it helps you calculate the force that that field exerts on another ball. Each part of the second ball feels a different field value. And Gauss’ Law is of no help for, say, cubes. $\endgroup$
    – G. Smith
    Commented May 26, 2020 at 22:22

1 Answer 1

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I wasn't sure if it is possible to say that $\overrightarrow{dF_{A\to B}}=Q_B\space dV\space\overrightarrow{E_A}$

It isn’t. This is dimensionally inconsistent: the left side has the dimensions of force but the right side has the dimensions of force times volume. The $Q_B$ should be $\rho_B$. You also don’t know what $\overrightarrow{E_A}$ is without integrating over all the charges in ball $A$.

However, you are thinking in the right direction. You can write the force between two extended objects as a six-dimensional integral that sums up the infinitesimal forces between each infinitesimal element of charge in one body and each infinitesimal element of charge in the other body. For two spherically-symmetric balls this integral is doable analytically but rather messy. (Try it numerically! Then try it for cubes instead of balls.)

An equivalent two-part approach like what you were attempting is to consider one body, do a three-dimensional integral to compute its field everywhere, and another three-dimensional integral to compute the force that this field exerts on all the charges in the other body.

But, in general, it is easier to work with integrals for potential energy or potential, because these involve scalars rather than vectors. For example, you can compute the potential inside and outside a spherical shell of uniform surface charge density by doing a fairly easy integral. Once you know this potential, another easy integral gives you the potential energy of two shells. From this you can derive the potential energy for two spherically-symmetric balls, by considering the balls to be made up of concentric shells. From the potential energy, you can derive the force between the balls.

And, lo and behold, if the balls don’t interpenetrate, the result will be that the spherically-symmetric balls behave as if they were simply point charges at their centers.

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  • $\begingroup$ Thank you for your answer, and yes that was supposed to be $\rho_B$, I missed when copying this out... $\endgroup$
    – alexisrdt
    Commented May 26, 2020 at 22:15

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