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

AI Thread Summary
The discussion revolves around solving a physics problem involving a mass hanging from a pendulum in an electric field. The initial approach of equating forces (tension, weight, and electric force) was confirmed as correct, while attempts to use conservation of energy were deemed inappropriate for this scenario. It was clarified that even when the net force is zero, the charge continues to move past the equilibrium point due to its kinetic energy. An analogy with a spring was provided to illustrate that a mass can overshoot its equilibrium position after being released. The conversation highlights the importance of understanding the dynamics of forces and motion in such problems.
spsch
Messages
111
Reaction score
21
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?
 

Attachments

  • IMG_20190622_150404.jpg
    IMG_20190622_150404.jpg
    42.5 KB · Views: 459
Physics news on Phys.org
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.)
 
  • Like
Likes spsch
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.)
 
  • Like
Likes spsch
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!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top