1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Vxy=Vy-Vx

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

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

    User Avatar

    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 ..

    Vxy=Vy-Vx

    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 !
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook