The Electric Field and Potential Energy in a Spring System

[Istiak]

Homework Statement

Electric field

Relevant Equations

F=kq/r^2

Solution attempt

 

[berkeman]

Find the electric field where? Can you post the full problem statement, and maybe the figure that is in the problem sheet that you are working from? Thanks.

 

[Delta2]

Seems to me you don't like to write a lot. Maybe english is not your mother language ok.

The attachment explains the problem setup but then it offers some choices but it doesn't mention what is the question. It looks like it is asking what is the total potential energy of the system.

And besides that, on your own work you seem to work towards finding the equilibrium position for charge q. Is that what you trying to find?

 

[Istiak]

Seems to me you don't like to write a lot. Maybe english is not your mother language ok.

The attachment explains the problem setup but then it offers some choices but it doesn't mention what is the question. It looks like it is asking what is the total potential energy of the system.

And besides that, on your own work you seem to work towards finding the equilibrium position for charge q. Is that what you trying to find?

Yes! It is. That's what I wrote. I was trying to find that value. Potential energy depends on the X²

 

[AF Fardin]

Yep

 

[Delta2]

Well I still don't understand your work.

At start you write $F=qE$ or $k''x=qE$
But after some point you write $F'+F=qE$
And later $F''+F'+F=qE$
only one case can be true, or I just don't understand something here.

 

[berkeman]

Yep

I was about to delete your reply as unnecessary, but when I checked your previous posting history, I see that you posted about a similar/same problem earlier today and the post was deleted because you showed no effort. Your attachment is added below; are you and @Istiakshovon in the same class?

 

[Steve4Physics]

Hi @Istiakshovon. I will also chip in.

“Title: Find the electric field”
The question is apparently not about finding the electric field!

“Homework Statement: Electric field”
That’s not a homework statement. You need to give the complete question.

Relevant Equations: F = kq/r^2
The equation is wrong. Maybe you mean E = kq/r^2 or F = kq₁q₂/r^2. However, neither of these equations is relevant to your question!

Like @Delta2, I too I can’t understand your solution attempt.

You have ignored @berkeman's post (#2). Is there a reason? I don’t think you will get help unless you at least post the complete question, as requested. (I can possibly guess what the question is but I might be wrong.)

 

[Istiak]

I was about to delete your reply as unnecessary, but when I checked your previous posting history, I see that you posted about a similar/same problem earlier today and the post was deleted because you showed no effort. Your attachment is added below; are you and @Istiakshovon in the same class?View attachment 284973

Actually, he was new to the community. He didn't understand how the community works. He wasn't familiar with these tools. So, he told me to create new thread here cause, I had talked about the community to him. Then, I had posted it. I hope you understood. I know that "Yep" comment is unnecessary. So, you can delete that. He doesn't know code of conduct of the community either. I haven't read COC also.

 

[Istiak]

Hi @Istiakshovon. I will also chip in.

“Title: Find the electric field”
The question is apparently not about finding the electric field!

“Homework Statement: Electric field”
That’s not a homework statement. You need to give the complete question.

Relevant Equations: F = kq/r^2
The equation is wrong. Maybe you mean E = kq/r^2 or F = kq₁q₂/r^2. However, neither of these equations is relevant to your question!

Like @Delta2, I too I can’t understand your solution attempt.

You have ignored @berkeman's post (#2). Is there a reason? I don’t think you will get help unless you at least post the complete question, as requested. (I can possibly guess what the question is but I might be wrong.)

As I mentioned in my previous reply that this question is not mine. It's another person's. I didn't have much more time to look at his works. So, that what I simply understood without reading answer and question, I had just added that. As I said I didn't have much more time, that's why I added unnecessary **Homework Statement**

 

[Istiak]

I was trying to edit my question. Unfortunately, there's no edit button. Is it bug? The thread was migrated from Calculus and Physics. I think that's why I don't have anymore access to edit the question.

 

[Delta2]

I was trying to edit my question. Unfortunately, there's no edit button. Is it bug?

No it isn't bug, the forum software doesn't let you edit the post a few hours after it was initially posted.

 

[Istiak]

No it isn't bug, the forum software doesn't let you edit the post a few hours after it was initially posted.

So, How can I edit Homework Statement?

 

[Delta2]

Just post a new message here with the full homework statement. Later when a moderator (for example @berkeman) might see your new message, he will copy paste the homework statement to the old message.

 

[Istiak]

Homework Statement: The second mass (m) also carries a charge q while the mass M is uncharged and made of an insulator. The masses are allowed to move only along the horizontal direction. The distance between the plates A and B as shown in the figure is fixed. Moreover there is a horizontal uniform electric field E that runs between A and B plates and points from left to right.

https://www.physicsforums.com/attachments/e4e2c5e8-5062-4bbf-94a7-55bbd56e393c-jpeg.284961/

 

[Delta2]

Homework Statement: The second mass (m) also carries a charge q while the mass M is uncharged and made of an insulator. The masses are allowed to move only along the horizontal direction. The distance between the plates A and B as shown in the figure is fixed. Moreover there is a horizontal uniform electric field E that runs between A and B plates and points from left to right.

https://www.physicsforums.com/attachments/e4e2c5e8-5062-4bbf-94a7-55bbd56e393c-jpeg.284961/

Ok fine for the description of the problem setup, but you didn't tell us what is the question asked. Is it "Find the equilibrium position for mass m"?

 

[Istiak]

Ok fine for the description of the problem setup, but you didn't tell us what is the question asked Is it "Find the equilibrium position for mass m"?

I am forwarding it to @AFFardin. @AFFardin Reply to him using the reply button to his reply

 

[AF Fardin]

Well I still don't understand your work.

At start you write $F=qE$ or $k''x=qE$
But after some point you write $F'+F=qE$
And later $F''+F'+F=qE$
only one case can be true, or I just don't understand something here.

F"+F'+F=qE
This is true I think !

 

[AF Fardin]

F"+F'+F=qE
This is true I think !

At First, I wasn't sure about it!
I was trying and thinking with some roughs!
And that's what made you confused. My final thought is this (F"+F'+F=qE).
Still don't know that my approach is correct or wrong!

 

[AF Fardin]

I was about to delete your reply as unnecessary, but when I checked your previous posting history, I see that you posted about a similar/same problem earlier today and the post was deleted because you showed no effort. Your attachment is added below; are you and @Istiakshovon in the same class?View attachment 284973

Yes

 

[AF Fardin]

Ok fine for the description of the problem setup, but you didn't tell us what is the question asked. Is it "Find the equilibrium position for mass m"?

[6/25, 09:49] AF_Fardin: Both X and Y
[6/25, 09:49] AF_Fardin: Means for both masses

 

[Delta2]

ok so you want the positions of the two masses for when the whole system is in equilibrium. Hold on while I work on my reply (might take a while cause I am a bit busy with other things at the moment.)

 

[Delta2]

Ok I think the correct system of equations is
$$Kx-K'y=0$$
$$K'y+Eq-K''z=0$$
$$x+y+z=L$$
The first equation is from the equilibrium condition on mass M, the second from the equilibrium condition on mass m, and the third is self explanatory

P.S We assume that the springs have zero natural length.

 

[AF Fardin]

Ok I think the correct system of equations is
$$Kx-K'y=0$$
$$K'y+Eq-K''z=0$$
$$x+y+z=L$$
The first equation is from the equilibrium condition on mass M, the second from the equilibrium condition on mass m, and the third is self explanatory

P.S We assume that the springs have zero natural length.

Wow thank you !
Could you please explain it explicitly for me?

 

[Istiak]

Ok I think the correct system of equations is
$$Kx-K'y=0$$
$$K'y+Eq-K''z=0$$
$$x+y+z=L$$
The first equation is from the equilibrium condition on mass M, the second from the equilibrium condition on mass m, and the third is self explanatory

P.S We assume that the springs have zero natural length.

But, actually your approximation doesn't match with any option. What should I do next (with this equation)?

 

[Delta2]

But, actually your approximation doesn't match with any option. What should I do next (with this equation)?

Could be that my equations are wrong. Can you tell me what are the available options, maybe i can do some reverse engineering of some sort ...

 

[Istiak]

Could be that my equations are wrong. Can you tell me what are the available options, maybe i can do some reverse engineering of some sort ...

Look at this picture..
1.

$$\frac{1}{2} kx^2 + \frac{1}{2} ky^2 - qEY$$

2.

$$\frac{1}{2} kx^2 + \frac{1}{2} ky^2 - qE(Y+X) + \frac{1}{2} K''(L-x-y)^2$$

3.

$$\frac{1}{2}kx^2+\frac{1}{2}ky^2-qEY+\frac{1}{2} (L-x-y)^2$$

4.

$$\frac{1}{2}kx^2+\frac{1}{2}ky^2-qE(x+y)$$

5.

$$\frac{1}{2}kx^2+\frac{1}{2}ky^2+\frac{1}{2}k''(L-x-y)^2$$

Unfortunately, MathJax isn't working. I had tried with "####" also. I had tried as we use in SE or, CD

 

[Delta2]

Ehm I don't think these options are candidates to the equilibrium position , they are rather candidates for the total potential energy of the system. Try option 2 and tell me if it is correct.

 

[Steve4Physics]

As I mentioned in my previous reply that this question is not mine. It's another person's. I didn't have much more time to look at his works. So, that what I simply understood without reading answer and question, I had just added that. As I said I didn't have much more time, that's why I added unnecessary **Homework Statement**

My apologies. You are just trying to help someone out. Well done for that!

But note, even after 28 posts, we do not know what the actual question is! If at all possible, post the complete question. We shouldn't have to make any assumptions such as "the springs have zero natural length" (@Delta , Post #23); all essential information should be included as part of the question.

 

[Istiak]

Ehm I don't think these options are candidates to the equilibrium position , they are rather candidates for the total potential energy of the system. Try option 2 and tell me if it is correct.

Actually, that was like an exam (Olympiad). My friend had joined to that. Then, he thought his work isn't correct that's why he told me to post it here.

And, no matter what I choose the site won't show correct answer. That's why I/he wants to know the approximation/answer.

 

[Istiak]

My apologies. You are just trying to help someone out. Well done for that!

But note, even after 28 posts, we do not know what the actual question is! If at all possible, post the complete question. We shouldn't have to make any assumptions such as "the springs have zero natural length" (@Delta , Post #23); all essential information should be included as part of the question.

Actually, I had gave the whole question and, figure in following picture. You can see the whole question in the image also.

 

[Delta2]

Actually, I had gave the whole question and, figure in following picture. You can see the whole question in the image also.

Actually I can't see what the question is. I can read only the description of the system. What I can guess is asking (judging from the available options) is "What is the total potential energy of the system?"

 

[Istiak]

Actually I can't see what the question is. I can read only the description of the system. What I can guess is asking (judging from the available options) is "What is the total potential energy of the system?"

Yes! You got the question..

 

[Delta2]

Yes! You got the question..

The correct option according to my opinion is option 2.

 

[Steve4Physics]

And, no matter what I choose the site won't show correct answer.

I'm not surprised!

@Delta has (I believe correctly) guessed that the question is 'What is the potential energy?'. But original question does not even mention potential energy!

The meaning of distances X and Y is very unclear. The question-wording does not mention X or Y. And the diagram in Post#1 does not show X and Y properly. (An accurate diagram showing the meanings of X and Y would help a lot.)

Assuming you have posted the full question, then it seems that whoever wrote the original question made a lot of bad mistakes/omissions. Without guesses/assumptions, the question is unanswerable.

 

[Istiak]

I'm not surprised!

@Delta has (I believe correctly) guessed that the question is 'What is the potential energy?'. But original question does not even mention potential energy!

The meaning of distances X and Y is very unclear. The question-wording does not mention X or Y. And the diagram in Post#1 does not show X and Y properly. (An accurate diagram showing the meanings of X and Y would help a lot.)

Assuming you have posted the full question, then it seems that whoever wrote the original question made a lot of bad mistakes/omissions. Without guesses/assumptions, the question is unanswerable.

Yes! That's true. I couldn't understand the question either. But, when I looked at option it was showing like they want us to find Potential Energy...

 

[kuruman]

The correct option according to my opinion is option 2.

I agree with your opinion and I also agree that the problem could have been stated better. This is what I think is the case. We start in zero electric field with the masses at equilibrium and the springs unstretched. Now we turn the field ON. Say the charged mass experiences an electric force to the right. Both masses will move to the right and we have the new equilibrium configuration as shown in the picture. The key question is how to interpret X and Y. The only interpretation that makes sense in view of the answers is that they represent the extra amount by which each spring is stretched as the charged mass moves to the right. Then one must add up the potential energy changes of the three springs plus the change in electric potential energy.

 

[Steve4Physics]

... We start in zero electric field with the masses at equilibrium and the springs unstretched. Now we turn the field ON. Say the charged mass experiences an electric force to the right. Both masses will move to the right and we have the new equilibrium configuration as shown in the picture. The key question is how to interpret X and Y. The only interpretation that makes sense in view of the answers is that they represent the extra amount by which each spring is stretched as the charged mass moves to the right. Then one must add up the potential energy changes of the three springs plus the change in electric potential energy.

That sounds like the intention of the question's original author.

But the spring on the right gets compressed by an amount X+Y (the same as the distance m moves, which is independent of L). The overall potential energy change (assuming q is positive) is then:
½kX² + ½k’Y² - qE(X+Y) + ½k’’(X+Y)²

This is not in the answer-list. I guess that the answer-list is also wrong!

It may be worth noting that (assuming no losses)
½kX² + ½k’Y² - qE(X+Y) + ½k’’(X+Y)² = 0
because
work done by electric field = gain in elastic potential energy.
or equivalently
loss of electrical potential energy = gain in elastic potential energy.

 

[kuruman]

That sounds like the intention of the question's original author.

But the spring on the right gets compressed by an amount X+Y (the same as the distance m moves, which is independent of L). The overall potential energy change (assuming q is positive) is then:
½kX² + ½k’Y² - qE(X+Y) + ½k’’(X+Y)²

This is not in the answer-list. I guess that the answer-list is also wrong!

It may be worth noting that (assuming no losses)
½kX² + ½k’Y² - qE(X+Y) + ½k’’(X+Y)² = 0
because
work done by electric field = gain in elastic potential energy.
or equivalently
loss of electrical potential energy = gain in elastic potential energy.

You are absolutely correct, good catch. I derived the same expression but didn't notice the difference between my expression and theirs. It seems that the question's author had one X & Y definition for the first two springs and another for the third spring.