Find the charge of a mass hanging from a pendulum in an electric field

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SUMMARY

The discussion focuses on calculating the charge of a mass suspended from a pendulum in an electric field using force equilibrium and conservation of energy principles. The correct approach involves equating the forces of tension, gravitational force (mg), and electric force (EQ), leading to the formula Q = (mgx)/(Ed). Participants clarify that while the electric field does work to elevate the charge, it does not stop at the equilibrium point due to kinetic energy, which allows it to swing past that point.

PREREQUISITES
  • Understanding of force equilibrium in physics
  • Familiarity with electric fields and forces (E, Q, d)
  • Knowledge of gravitational potential energy (mgx)
  • Basic concepts of kinetic energy and motion
NEXT STEPS
  • Study the principles of force equilibrium in electric fields
  • Learn about gravitational potential energy and its calculations
  • Explore the dynamics of pendulum motion in electric fields
  • Investigate the relationship between kinetic energy and oscillatory motion
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Physics students, educators, and anyone interested in understanding the dynamics of charged particles in electric fields and their motion in pendulum systems.

spsch
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Homework Statement
A charge with mass 1 gram hanging from a pendulum is at equilibrium 12 cm above the lowest vertical position. E= 9500. l the length of the pendulum is 55 cm.
Relevant Equations
QEd = mgh?
Hi, so I was able to solve this problem by just equating the forces (Tension, mg, and EQ).

But I thought I could also solve this problem with Conservation of Energy.
However, I calculated it several times, and I never get the right answer this way.
Doesn't the Electric Field do the work to put this charge at its new Gravitational Potential Energy position 12cm higher?
Or the -difference in Electric Potential is the Gravitational Potential gained?

## Q*E*d = m*g*x ## and therefore ## \frac {mgx}{Ed} = Q ## ?
Or is there another Energy term I am missing?
 

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spsch said:
Hi, so I was able to solve this problem by just equating the forces (Tension, mg, and EQ).
That is the correct approach.

spsch said:
But I thought I could also solve this problem with Conservation of Energy.
No, this is a force problem, not a conservation of energy problem. (If you released the charge from rest at the bottom position, it would swing up past the equilibrium point.)
 
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Hello @Doc Al , Thank you very much for answering!
The correct approach has been the most obvious. I'm trying to do old problems in several ways now to practice.

I'm sorry I don't understand yet. Why would the charge swing past the equilibrium point?

Maybe I'm picturing it wrong.
I imagined the charge being released in the middle and the electric field pushed it up to the current position.
Until the forces cancel each other.
x (12cm) above the original position and a distance d along the electric field.
The electric field accelerates the charge but is losing magnitude (## cos(θ) ##) gravity (increasing in magnitude with (## sin(θ) ##)) is decelerating the charge until it comes to rest?

I'm sorry I'm sure I'm being super difficult!
 
spsch said:
Why would the charge swing past the equilibrium point?
Just because the net force is zero doesn't mean it stops.
spsch said:
I imagined the charge being released in the middle and the electric field pushed it up to the current position.
Until the forces cancel each other.
At that point, it's still moving. So it will keep going until it runs out of kinetic energy.

Here's another example: Say you had a spring. You attach a mass and let it drop. The mass drops, but it keeps moving past the equilibrium point. It doesn't just stop at that point. (In the case of the spring, it oscillates.)
 
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Doc Al said:
Just because the net force is zero doesn't mean it stops.

At that point, it's still moving. So it will keep going until it runs out of kinetic energy.

Here's another example: Say you had a spring. You attach a mass and let it drop. The mass drops, but it keeps moving past the equilibrium point. It doesn't just stop at that point. (In the case of the spring, it oscillates.)
@Doc Al Thank you. The spring did it for me!
 

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