Find the electric field

  • #36
Istiak
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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...
 
  • #37
kuruman
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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.
 
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  • #38
Steve4Physics
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... 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.
 
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  • #39
kuruman
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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.
 
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