Part B has the same setup as Part A, so
we'll use the same diagram. Except now, we
know the distance between the particles.
We want to find the charge of the
positive particle, so we want to find qₚ.
Let's pretend that there is a pin
keeping particle B in place, so we can
look at the forces acting on A first.
There is a force pulling particle A
towards the positive particle, and there
is a force pushing particle A away from
particle B. Now for particle A to remain
stationary, it must have a net force of
zero acting upon it. F_B pushes A to the
left while F_P pulls a to the right. So a
net force of zero is certainly possible.
Now we want F_B to equal F_P. That way,
A experiences no net force and thus
remains stationary. The distance between
particle A and the positive particle is
smaller than the distance between
particle A and particle B. In other words,
dₐ is smaller than dₜ. So for the forces
A experiences from these particles
individually to be the same, then
particle B must have a bigger charge
than the positive particle to compensate
for the differences in distance. Using
Coulomb's law again we get these two
equations. Equating them and cancelling
terms, putting in the numbers, and then
solving for qₚ, we get a negative number,
which is telling us that the charge the
positive particle must have is negative.
So it is in fact a negative particle;
that's weird.
However, we can make sense of this by
remembering that force is a vector, and
so it has direction. F_B points in the
opposite direction to F_P. Since we want
them to point in opposite directions but
be equally strong, we actually wanted F_B
equals negative F_P. If we carry this
through the working out, we'll find that
the charge the positive particle has is
in fact +0.4 C.
Remember that due to the difference in
distance, the positive particle must have
a smaller charge than the negative particle.
0.4 C is smaller than 3 C, so this condition is satisfied,
and our answer is plausible. Now, if we
run through this whole process again
with particle A fixed and try to balance
the forces on particle B, we'll actually
get the same answer for the charge of
the positive particle. It isn't
immediately obvious why this should be
the case, but if we dig a bit deeper, we
can find out why it must be so. So let's
look at our system again. Remember how in
Part A we changed the distance such that
the positive particle experience is the
same amount of force from both A and B.
The same amount of force pulls particle
A towards the positive particle, as
discussed in previous videos. This is the
same for the interaction between the
positive particle and particle B.
Particle B exerts a force on particle A,
pushing it to the left. Let's call this
force F_C. Particle A also exerts the
same amount of force on particle B. In
Part B of the question, we were trying to
find the value qₚ must take such that F_A
equals F_C. This gives us a net force
of zero acting on particle A. But at the
same time, it also gives a net force of
zero acting on particle B. So for Part B,
the charge that the positive particle
has, which would keep particle A
stationary, would also keep particle B
stationary.
