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Basic electric force/field question I can't get

  • Thread starter Kuzon
  • Start date
  • #1
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Homework Statement


A particle has a charge of +1.7 μC and moves from point A to point B, a distance of 0.26 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA - EPEB = +8.1 x 10-4 J. (a) Find the magnitude of the electric force that acts on the particle. (b) Find the magnitude of the electric field that the particle experiences.

Homework Equations


F= qV/d

The Attempt at a Solution


What's wrong with this reasoning:
F = V/d and V=Ed so E=F/q=V/d
Therefore F=qV/d
substitute all those values into that formula to get the electrostatic force and it's wrong?

I also tried b) using E=kq/d^2 and that didn't work either! (I got 226331.3609 as the answer)

By the way, there are no rounding requirements or anything, only a 2% margin of error so it's not that.

Thanks a lot for your help in advance!
 

Answers and Replies

  • #2
blue_leaf77
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What you are given is ##\Delta E_p## which is equal to ##q\Delta V##. By the way, what's the correct answer?
 
  • #3
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What you are given is ##\Delta E_p## which is equal to ##q\Delta V##. By the way, what's the correct answer?
Right so I tried going:
deltaE/q = deltaV
Then using F=(q*deltaV)/d to get the electrostatic force and I got the wrong answer again
 
  • #4
blue_leaf77
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What's the correct answer according to your source?
 
  • #5
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What's the correct answer according to your source?

I don't know sorry
 
  • #6
blue_leaf77
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Then how did you know that your answer was wrong? Is it a homework given to you?
 
  • #7
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Then how did you know that your answer was wrong? Is it a homework given to you?

It's this homework website and you get 5 attempts at getting the correct answer, it tells you when it is wrong but doesn't give you hints or the correct answer.
 
  • #8
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Could someone please help with this question? Really struggling
 
  • #9
blue_leaf77
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Can you post your own answer?
 
  • #10
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Can you post your own answer?

I did it's in the OP. Do you know how to solve the question?
 
  • #11
Charles Link
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They give you the change in potential energy ## \Delta W=F \cdot d=q V ##. The change in voltage is often also referred to as the change in potential, but here they were quite specific that the particle experienced a change in its "potential energy", so that ## \Delta W=q V ##. (using ## W ## here for energy to save the letter ## E ## for the electric field. Notice also that ## V ## stands for voltage in electrical problems, and only becomes energy when multiplied by the charge ## q ##.)
 
Last edited:
  • #12
NascentOxygen
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What's wrong with this reasoning:
F = V/d and V=Ed so E=F/q=V/d
Therefore F=qV/d
substitute all those values into that formula to get the electrostatic force and it's wrong?
You start with F=V/d and show that F=qV/d? Something is wrong there. The units on each side of F=V/d don't match, so it can't be right.

What value did you use for V?
 
  • #13
blue_leaf77
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I did it's in the OP. Do you know how to solve the question?
You mean this: 226331.3609 ? Isn't that your answer for (b). I asked for your answer for part (a).
 
  • #14
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They give you the change in potential energy ## \Delta W=F \cdot d=q V ##. The change in voltage is often also referred to as the change in potential, but here they were quite specific that the particle experienced a change in its "potential energy", so that ## \Delta W=q V ##. (using ## W ## here for energy to save the letter ## E ## for the electric field. Notice also that ## V ## stands for voltage in electrical problems, and only becomes energy when multiplied by the charge ## q ##.)


Thanks so much for this, I got it now.
 

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