Solving Electromag Homework: Can You Help?

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The discussion revolves around solving an electromagnetism homework problem involving two fixed charged beads and a third movable bead. The user initially believes their calculations for the equilibrium position of the third charge are correct but receives an error from their program. They derive the equation for forces acting on the third charge and simplify it to find the position x in terms of d. Ultimately, the user discovers that the program's rejection of their answer was due to a formatting issue, specifically the absence of a leading zero. The correct approach to solving the problem involves ensuring proper notation and understanding the equilibrium condition where forces from both charges balance.
notmetalenough
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I'm doing my first electromag homework so there will probably be a few more of these.

Two small beads having positive charges 19q and q are fixed at the opposite ends of a horizontal insulating rod extending from the origin (the location of the larger charge) to the point x = d. As in the figure below, a third small charged bead is free to slide on the rod.

So I figure that

F = ((k * 19q Q)/ x^2) - (k* q * Q)/(d-x)^2)

so 19/x^2 = 1/(d-x)^2

d-x = x / sqrt(19)

d = x + x/sqrt(19)

= x * (1+sqrt(19))/sqrt(19)

so x = d * (sqrt(19))/(1 + sqrt(19))

which would be about 0.81339

apparently it's the wrong answer. Can anyone tell me why, or where I went wrong?
 
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Are you trying to find the point of equilibrium for the third charge? If so, it will find equilibrium at the point where the force from both charges cnacel each other out (equal each other) or when the electric field is zero. These occur at only one point and they are the same point.
 
right, but i have to find it as a ratio of d.

i can't find anything wrong with how i set it up, I made the charges proportional, and equal to each other and then solved for x, my answer should be sound, but the program won't accept it.
 
okay, false alarm. I did have the right answer, my program wouldn't accept it without the leading 0 and wouldn't tell me that that was the reason I was getting it wrong.
 
((k * 19q Q)/ x^2) = (k* q * Q)/(d-x)^2) is where you should start, solving for x.
 
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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