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What is the work needed to move a charge

  1. Feb 12, 2015 #1
    1. The problem statement, all variables and given/known data
    3 charges are spread as drawn. the first question was what's the field at point B, but that i solved.
    The second question was what is the work needed to bring the charge q that's at point A to point B.

    2. Relevant equations
    The potential at a point: ##V=k\frac{q}{r}##
    The work: ##W=q\cdot \Delta V##

    3. The attempt at a solution
    I don't think i have to take into consideration the charge that i should move. it contributes to the field at B but it changes. and i am not sure if i have to take him in consideration also at the start point A because r=0 and V→∞ so:
    $$V_A=kq\left(\frac{1}{3}+\frac{4}{4}\right)=\frac{4}{3}kq$$
    $$V_B=kq\left(\frac{1}{\sqrt{13}}+\frac{4}{\sqrt{20}}\right)$$
    $$W=kq^2\left(\frac{1}{\sqrt{13}}+\frac{4}{\sqrt{20}}-\frac{4}{3}\right)$$
    I think it should be:
    $$W=kq^2\frac{1}{6}$$
     

    Attached Files:

  2. jcsd
  3. Feb 12, 2015 #2

    gneill

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    Hint: What's the total electric potential energy of the system before, and then after the move?
     
  4. Feb 12, 2015 #3

    BvU

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    With VA you mean the potential at point B and with VB you mean the potential at point A ?

    And with 1/6 you mean the outcome of the calculation of ##\left(\frac{1}{\sqrt{13}}+\frac{4}{\sqrt{20}}-\frac{4}{3}\right) ##?
     
  5. Feb 12, 2015 #4
    To BvU: yes to all.
    To gneil: the difference in potential energy between the first and second positions equals my calculation with the work and is ##\Delta E=0.162\cdot K\cdot q^2##
    So the answer 1/6kq2 isn't correct?
     
  6. Feb 12, 2015 #5

    gneill

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    No, not correct. You don't want the potential at a given location, you want the total energy of the system (the energy required to assemble the system from scratch in a given configuration).
     
  7. Feb 12, 2015 #6
    Yes, that's what i did and what i meant, i assembled both configurations and got the difference in potential energy of the whole system
     
  8. Feb 12, 2015 #7

    gneill

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    I'm not sure what you did precisely since you haven't shown all your work. Also I'm a bit confused as to what the expressions you've presented really pertain to (If they really are what you say they are, i.e., no errors snuck in).

    What is the expression you obtained for the energy of the initial system configuration?

    What is the expression you obtained for the energy of the final configuration?
     
  9. Feb 12, 2015 #8
    I made only the differences between the initial and final states.
    I first bring charge q to point C from infinity, WC=0. then i bring charge q to places A or B, the difference between them is the difference between the hypotenuse (point A) and the edge of triangle ABC:
    $$\Delta V_{B-A}=kq\left(\frac{1}{3}-\frac{1}{\sqrt{13}}\right)$$
    Then i bring chrge 4q to position D. the only difference is from the charge at positions B and A since the charge at position C is the same in the initial and final states:
    $$\Delta V_D=kq\left(\frac{1}{4}-\frac{1}{\sqrt{20}}\right)$$
    And the total Energy difference for the initial and final states:
    $$\Delta E=kq^2\left[\left(\frac{1}{3}-\frac{1}{\sqrt{13}}\right)+4\left(\frac{1}{4}-\frac{1}{\sqrt{20}}\right)\right]=0.162\cdot kq^2$$
     

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  10. Feb 12, 2015 #9

    gneill

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    That looks better. What units do you associate with the numerical value 0.162 ? Remember, the result should be energy (Joules).
     
  11. Feb 12, 2015 #10
    The units are ##\frac{1}{m}## since that's how i made the 0.162: it comes from ##\frac{1}{r}## and it also fits to the final solution to get energy because the units of k and q are:
    $$0.162\left[\frac{1}{m}\right]\cdot k\left[\frac{N\cdot m^2}{Coulon^2}\right]\cdot q[Coulon^2]=[N\cdot m]$$
     
  12. Feb 12, 2015 #11

    gneill

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    Yup. Excellent.
     
  13. Feb 12, 2015 #12

    BvU

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    Well done. Only reason (well, two reasons) I posted was I found the naming slightly confusing and the last step a bit weird:
    where W changes sign and assumes an unearned appearance of exactitude.

    To Karol: you did just fine. Gneill: Karol was doing the right thing. Your post #2 and #5 look as if you didn't agree ?
     
    Last edited: Feb 12, 2015
  14. Feb 12, 2015 #13
    Indeed it's strange that the work in post #1 is negative. the potential at B should be higher than at A since the charges are closer and thus more energy is needed to bring the charge to there. and i think that positive work means that you have to invest work, no?
    And why, if the two solutions are equivalent, the work in the later is positive, meaning the energy of the combined system is indeed higher?
     
  15. Feb 13, 2015 #14

    BvU

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    If the potential at point A is VA, the work per unit charge needed to bring a charge from infinity to point A is ##V_A - V_\infty##
    If the potential at point B is VB, the work per unit charge needed to bring a charge from infinity to point B is ##V_B - V_\infty##
    Then the work per unit charge needed to bring a charge from point A to point B is ##V_B - V_A##

    You confuse me with your naming:
    That way you need to calculate your VA ##-## your VB. Your VA(4/3 kq) is more positive than your VB, so W is positive. As you thought it should be. Apparently you confused yourself too, but I didn't manage to bring that across clearly in my post #3.

    [edit] work ##\rightarrow## work per unit charge
     
    Last edited: Feb 13, 2015
  16. Feb 13, 2015 #15
    I thank you very much all of you
     
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