Asked by Pat

2 point charges with the same magnitude (5*10^-8) but opposite signs are 10m apart. What is the magnitude and direction of the force at the midpoint between the charges?

I am supposed to do E=(k*q1)/r
is r (distance) 10m or 5m (midpoint)
Answered by bobpursley
well, the forces are toward the other charge.

There is NO FORCE at any point unless there is a charge there. There is an E field, yes, but that is not force. If you put any charge there, there would be a force away from one of the original charges, and toward the other.
Answered by Pat
Let's say there was a point charge at the midpoint. The question I was told was kinda confusing and I'm going off what I can remember. Point charges are normally protons.

It would be close to the electron; what would I write as the equation?
Answered by Anonymous
E=(k*q1)/r
No,
E=(k*q1)/r^2

E is not a force. It is the strength of the electric field. If you put another charge, q2 at that point, you get a force on it equal to g2*E
Answered by Anonymous
by the way, the potential energy per unit charge, call it V or voltage, is like the equation you typed and is proportional to 1/r

if you integrate the E field of a charge in from infinity to the point at r from the charge, you get your type of equation but it is voltage, not force.
Answered by Pat
g2*E = gravity2 times E?

or is that q2*E
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