Force on a Negative Charge Q1 in the Presence of a Long Rod

AI Thread Summary
The discussion revolves around calculating the forces acting on a negative charge Q1 near a long, negatively charged rod. For Q1 at rest, the force is determined using Coulomb's law, yielding a force directed towards the rod. When Q1 moves out of the page, the force is zero due to the angle being 90 degrees, indicating no effective interaction. If Q1 moves leftward, the discussion suggests additional complexities may arise, prompting questions about whether electrostatic forces are the only factors to consider. The need for clarification on the forces involved highlights the intricacies of the scenario.
jaded18
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Consider a very, very long rod (length L) with total negative charge Q0,
uniformly spread out. The rod is located in the plane of the page, is parallel to the top of the
page, and moves right‐ward with a speed v. There is a negative charge Q1 located a small
distance D away from the rod. Neglect gravity.

a) What is the total force (magnitude and direction) acting on the charge Q1 if it is NOT
moving?
b) What is the total force (magnitude and direction) acting on the charge Q1 if it is moving
with out of the page at speed v1?
c) What is the total force (magnitude and direction) acting on the charge Q1 if it is moving
left‐ward at speed v1?

attempt at ANS:
a) coulomb's law = F = k(QQ)/(r^2) = k(Q_e)^2/(D^2)
b) 0 because angle=90?
c) ...

I need help.
 
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Are you sure that the only force in play here is the electrostatic force?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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