Electrostatics: A square with point charges on the corners

AI Thread Summary
In the discussion on electrostatics involving a square with point charges at each corner, the main problem is determining the position and charge of a fifth point charge needed for equilibrium. The solution requires placing the fifth charge at the center of the square with a charge of -0.957q. Participants emphasize the importance of using vector analysis to account for the forces acting on the fifth charge due to the four corner charges. The symmetry of the arrangement simplifies calculations, allowing for effective force balance. Understanding these principles is crucial for solving similar electrostatic problems.
fara0815
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Hello,
I am doing electrostatics at the moment and have difficulties to solve the following problem. Any hint that helps me to find the answer will be appreciated!

"On every corner of a square are movable point charges with the charge of 'q'. Where does a fifth movable point charge have to be and what charge does it need to have so that the system is in equilibrium?"

The answer is -0.957q and I do not know how to get that.
 
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Well, "the system needs to be in equilibrium"...What equation do you use to express the equilibrium of the system ?

marlon
 
Mh, my idea is that you can consider the four charges as one since the electric field lines between them equal out, so that only the lines on the outside of the square act on a point charge.
The force caused by the four charges that acts on the fifth charge has to be a great as the force that is caused by the fifth charge and acts on the four charges.
Is that what you mean?
 
I figured it out!
I tried to do it without vectors and that just does not work ;)

If you do it with vectors and since it is symmetrical, you can calculate the forces acting on on point charge. The fifth point charge needs to be in the center.
 
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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