Where does the electric force come from if an electron has no definite location? Say electron A is nearby another electron (B), so that they may repel each other. Electron B is in a position eigenstate (so it has a definite position). But electron A is not. How does electron A affect the acceleration of electron B? Does it "divide up" its electromagnetic force as if it were a charged object spanning the space that the wave function occupies, whose charge density is proportional to the value of the probability density function? Otherwise, how can electron B decide where to move?
Simply: if an electron can be in "multiple places at once", and the force it produces depends on its location, which location is "chosen" for that force?
...I know that $\exists$ a whole theory on this, Quantum Electrodynamics (thanks Feynman!!!), but I have not studied it. I have only ever taken an intro QM class as an undergraduate.
Edit: If the position eigenstate causes problems, let B be in an arbitrary eigenstate as well. The question is rephrased: if the positions are indeterminate, how is the force, which depends on them, calculated?
 A: Note that the problem you pose is non-realistic. If at a certain moment B is in a position eigenstate, $\delta (\vec r)$, at an extremely short time after , B can be everywhere is the universe with equal probability. You will see the effect of this, below.
But let's first calculate the force $<\vec F>$. In QM, the influence of between A and B goes as follows: let $\psi_A(\vec r)$ be the wave-function of the electron A, where the vector $\vec r$ connects A, wherever A is, with B.
Then the force of interaction is
$\vec F(\vec r) = -\frac {e^2 \vec r}{4 \pi \epsilon_0 |r|^2}$.
The average force between the two electrons is
$<\vec F> = \int d\vec r \int d\vec r' \psi_A^* (\vec r)\delta (\vec r') \frac {e^2 (\vec r - \vec r')}{4 \pi \epsilon_0 |\vec r - \vec r'|^3} \psi_A (\vec r) \delta (\vec r')$.
$=\int d\vec r d\vec r' \delta (0) |\psi_A (\vec r)|^2 \frac {e^2 \vec r}{4 \pi \epsilon_0 |\vec r|^2} = \delta (0)\int d\vec r |\psi_A^* (\vec r)|^2 \frac {e^2 \vec r}{4 \pi \epsilon_0 |\vec r|^2}$
So, we have a problem because the function $\delta (\vec r')$ has infinite norm. On the other side, if $\psi_A$ is spherically symmetrical, one gets $<\vec F> = 0$. 
For the case that $\psi_A$ is not spherically symmetrical we have to replace the wave-function of B by another function, let's name it $\psi_B (r')$, highly localized around the point $\vec r' = 0$, but normalized. In that case
$<\vec F>=\int d\vec r d\vec r' |\psi^* _B (\vec r')|^2 |\psi_A^* (\vec r)|^2 \frac {e^2 (\vec r - \vec r')}{4 \pi \epsilon_0 |\vec r - \vec r'|^3}$,
and since $\psi_B (r')$ is highly localized around $\vec r' = 0$ we can approximate,
$<\vec F>=\int d\vec r d\vec r' |\psi^* _B (\vec r')|^2 |\psi_A^* (\vec r)|^2 \frac {e^2 (\vec r)}{4 \pi \epsilon_0 |\vec r|^3} = \int d\vec r |\psi_A^* (\vec r)|^2 \frac {e^2 (\vec r)}{4 \pi \epsilon_0 |\vec r|^3}$.
Now I return to the next moment after localization. The function $\psi_B (\vec r')$ will be practically zero, everywhere. So, in the before last equation we will get $<\vec F> = 0$.
A: If your charged particle is not in a position eigenstate you can always write the position as a superposition of positional eigenstates. Therefore you would have a quantum superposition of forces on a test particle (weighted by the probability amplitudes).
For instance suppose you have a negative particle which is initially has the wavefunction
$$\psi^- = \delta(0),$$
i.e. is in the definite position state located at the origin. Now suppose you have another positive charge in the superposition state
$$\psi^+ = \left(\delta(-x) + \delta(+x)\right)/\sqrt{2}.$$
Now a short time later both particles will spread out a little, but the negative particle will be in a superposition of moving to the left and right, i.e.
$$\psi^-\approx\left(\delta(-\Delta x) + \delta(+\Delta x)\right)/\sqrt{2}.$$
Of course the positive charge will also change position (and spread out).
If we plotted this, the wavefunctions might look like the picture below.

If you tried to measure the position (of either or both particles), then you would get a wavefunction collapse and particles would both be on either the right or left (with 50% probability for each case). 
