1
$\begingroup$

The presentation of Coulomb's Law in various books occasionally has a note that the test charge, q2, must be small enough that it doesn"t alter the field of the first charge, q1. The same limitation applies to a test charge in any other stationary electric field. If those are the facts of life, then fine, but how can anyone confidently calculate the force between two macroscopic bodies, each with its own charge distribution? What are reasonable limits on charge level, physical size, and separating distance? It seems that one could correctly perform any and all required volume integrations and occasionally, if not often, produce invalid results. Moreover, for actual physical charged bodies, won't the electric field(s) alter the charge distribution on those material bodies? Is the solution to these issues simply some non-linear formulation that I just haven't seen yet?

$\endgroup$
  • $\begingroup$ If you have two bodies with a certain charge distribution, the net force that each body acts over itself is equal to zero due to Newton's third law. In this kind of problems it is usually considered that the charge distributions are fixed (i.e. the bodies are insulators), but you could solve the problem computationally applying the laws you already know. $\endgroup$ – TheAverageHijano Feb 21 at 8:08
  • $\begingroup$ I guess I asked two questions at once. Sorry. My basic question is, since Coulomb's Law requires the test charge to be small enough that it doesn't change the electric field from the other charge, how can that Law be of practical use when calculating force values? Also, any formulations derived from Coulomb's Law have the same limitation. $\endgroup$ – Classical Mechanics Guy Feb 22 at 2:35
2
$\begingroup$

Some of them could be complicated, if they don't have symmetry. The good starting point is by assuming symmetry. For example, Gauss' law can be used to solve for the electric field due to a sphere, plane charge density, cylinder etc.

These derivations use Coulomb's law as a starting point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.