Electric field at origin, zero or nonzero?

AI Thread Summary
The electric field at the origin, located between two point charges +q and -q, is nonzero due to their opposite polarities. The electric field vectors from each charge point in opposite directions, with the positive charge creating a field directed away from it and the negative charge creating a field directed toward it. At the origin, these fields do not cancel out; instead, the net electric field points from the positive charge toward the negative charge. If a charge of -q were placed at the origin, it would experience a force directed toward the positive charge, indicating the presence of a net electric field. Thus, the electric field at the origin is indeed nonzero and directed from +q to -q.
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Homework Statement



Point charges +q and -q are placed at +50 cm and -50 cm on the x-axis, respectively. Is the electric field at the origin (halfway between the charges) zero or nonzero? If nonzero, which direction does it point?

Homework Equations





The Attempt at a Solution



If they were like charges, both positive, then I would say the electric field would be 0. I'm not sure because they are opposite polarities. Please help.
 
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Try to think of the electric field lines; if you drew them what would they look like?
 
What would happen if I were to place a charge of -q at the origin?
How about +q? Would you agree that they would move? What does
this say?
 
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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