Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Very simply find potential where its zero question

  1. Aug 5, 2010 #1
    Very simply "find potential where its zero" question

    1. The problem statement, all variables and given/known data

    A +3.0 nC char age is at x= 0cm and a -1 nC charge is at x= 4cm. At what point or points on the axis is the electric potential zero?


    2. Relevant equations

    V= [tex]\frac{kQ}{r}[/tex]

    3. The attempt at a solution

    i have solved the question, first i tried two places, between the two charges, and after the second charge.

    1op3c0.jpg

    i solved for x= 3cm and x= 6cm.

    now the problem is what is i want to check in the negative zone?

    2qxasmp.jpg

    now is in
    the formula v1+v2 =0

    for r1, do i use r1= -p and r2 = 4-p, i am confused about doing it in the negative zone, do i just use the magnitude, if only magnitude, it would be r1=p and r2=4+p, which do i use?
     
  2. jcsd
  3. Aug 5, 2010 #2

    thrill3rnit3

    User Avatar
    Gold Member

    Re: Very simply "find potential where its zero" question

    Other than the underlined [ corrected ] part, both are actually the same. You just multiplied both sides by -1.

    So you're fine either way.
     
  4. Aug 5, 2010 #3
    Re: Very simply "find potential where its zero" question

    i did it, r1=p and r2= 4+p, i get p=6cm, now when i go back and plug in p=6 in the equation it does not satisfy, but =-6cm works, but the answer from solving the equation is p=6cm not p=-6cm, i am confused, help please.


    0 = [tex]\frac{k(3nC)}{P}[/tex] + [tex]\frac{k(-1nC)}{4+P}[/tex]
     
  5. Aug 6, 2010 #4

    thrill3rnit3

    User Avatar
    Gold Member

    Re: Very simply "find potential where its zero" question

    I solved for p using the above equation and got p= -6

    Maybe something wrong with your algebra?

    even with r1= -p and r2= -4-p, I still get p= -6
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook