Calculating charge dissipation given two hanging masses

  • Thread starter Thread starter noleguy33
  • Start date Start date
  • Tags Tags
    Charge Dissipation
AI Thread Summary
The discussion focuses on calculating the rate of charge dissipation for two hanging masses with identical charges that create an angle theta from the vertical. The user initially derived theta but struggled with the next steps, proposing a relationship between charge change and distance. Another participant clarified that the velocity should account for both horizontal and vertical components, suggesting the use of angular velocity. They recommended deriving the equation for theta with respect to time to relate it to the charge dissipation rate. This approach aims to simplify finding the desired rate of charge loss.
noleguy33
Messages
25
Reaction score
0

Homework Statement



Two hanging masses have an identical charge and create an angle theta from the vertical. If they start losing their charge and have an instantaneous velocity of XX m/s(given), find the rate in which they are losing the charge.

Homework Equations



Coulumbs Law

The Attempt at a Solution



The first part of the question was to find theta, which I did. I now have no idea how to handle it from here. The only thing I came up with is-

dq/dt = dx/dt * dq/dx

Where dq/dt is what I am trying to find, dx/dt is the given velocity and dq/dx would be the change in charge in terms of the distance from the vertical. I'm not sure how to find that.

Thanks.
 
Physics news on Phys.org
I don't think dx/dt is the instantaneous velocity, because the velocity should also include dy/dt. Instead, the charges are tracing out a circle, so their velocity should be L*d(theta)/dt, where L is the length of the string.

Since you already have an equation for theta, presumably in terms of q, why not derive both sides with respect to time? On the left side, you'll get d(theta)/dt. On the right side, you'll get a factor of dq/dt, which is the factor you're trying to solve for.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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}...
Back
Top