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Electric potential

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Derive an expression for p.d , V between point X and point Y in an electric field due to a single point charge +Q as shown in the figure attached .

    2. Relevant equations

    3. The attempt at a solution


    [tex]=\frac{Q}{4\pi\epsilon_o r_y}-\frac{Q}{4\pi\epsilon_o r_x}[/tex]


    I guess it's not that simple , because this question carries 5 marks .

    sorry , i am not sure which is the icon to wrap these math codes .. could someone pls modify my post ?
    Last edited by a moderator: Feb 27, 2010
  2. jcsd
  3. Feb 27, 2010 #2

    Doc Al

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    Staff: Mentor

    Looks OK to me. Were you supposed to derive it from the electric field?

    To use Latex, wrap it with tex tags:
    [tag]x^2[/tag] = [tex]x^2[/tex]

    replace 'tag' with 'tex'
  4. Feb 27, 2010 #3
    thanks , but this is another way of doing it ..


    and from work definition , dW=F dr

    [tex]V_{xy}=\frac{F dr}{q}-\frac{F dr}{q}[/tex]

    [tex]=\frac{1}{q}\int^{r_y}_{\infty} F dr-\frac{1}{q}\int^{r_x}_{\infty} F dr[/tex]

    [tex]=\frac{1}{q}\int^{r_y}_{r_x} \frac{Qq}{4\pi \epsilon_o r^2}[/tex]

    [tex]=\frac{Q}{4\pi \epsilon_o }[-\frac{1}{r}]^{r_y}_{r_x}[/tex]

    [tex]=\frac{Q}{4\pi \epsilon_o }(\frac{1}{r_x}-\frac{1}{r_y})[/tex]

    When it's derived from the definition of electric potentials , the final products are different , or can i say since electric potential is a scalar , so that shouldn't matter ?
  5. Feb 27, 2010 #4

    Doc Al

    User Avatar

    Staff: Mentor

    You need to calculate the work done per unit charge against the electric force, so you're missing a minus sign in the expression for force. (Signs matter!)
  6. Feb 27, 2010 #5
    oh , thank you very much sir !
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