Eventually, I will ask the question, "Why?".

So, I was reviewing basic electrostatics and something really dumb popped into my head.  It has to do with electric fields  and electrostatic forces on charged particles.  Here's some background for context.


If you have two or more charged particles, you can calculate the (vector) electric field they create.  The field will (and must) include the influence of the two (or more) charged particles - multiple charges will cause field line bending, etc.  

E(x,y,z) = SUM(k*Qi/r2 * rhat)

Using this you can calculate the field in infinite detail.

If you have done any of this work, you also know the formula:

F = E * Q

Where the force (F) on a charged particle (Q) is proportional to the electric field (E).

So, cutting to the chase, here it comes -

The thing is, if you want to calculate the force between two charged particles, the previously calculated electric field (E(x,y,z)) is useless, even though it included the contributions of all charged particles,.  You have to calculate the force caused by one particle on the other directly using:

F = k*Q1Q2/r2 * rhat


Why?

Why bother with the field if it is not useful for calculating forces?  Why doesn't F = E*Q work?

Hmm,... not really,... yes,... and no.  

The issue is the fields versus the forces.  You need to include all charges to define the electric field but also the need to ignore the field (or eliminate one charge from the field calculations) to find the forces.

If I include all charged particles, you can use E = Kq1/r2 to calculate (map) the electric field in its entirety over and region of space I care to bother with.  You can map right up to the charges themselves, then map on the other side, off to the left or right, etc.  But that E field map is not useful to then calculate the forces.  To get the forces, you start all over again, and calculate F = kq1q2/r2.

You might think you can calculate the vector E field, then multiply by q2 to get the vector Force .  After all, E = kq1/r2 and F = kq1q2/r2 (differing only by a factor of q2, a scalar quantity).  But you'd be wrong.



I did say it was a dumb question.

Superposition still works.
One at a time.

?