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Again an electric potential question ?

  1. Sep 23, 2009 #1
    Again an electric potential question ?????

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


    Question # 1

    A charge of 7.034 nC is uniformly distributed
    along the x-axis from −4 m to 4 m.
    What is the electric potential (relative to
    zero at infinity) of the point at 5 m on the
    x-axis?





    2. Relevant equations



    3. The attempt at a solution


    I got 21.0728 and its wrong :cry: How come ??
     
  2. jcsd
  3. Sep 23, 2009 #2
    Re: Again an electric potential question ?????

    how did you get this answer?
     
  4. Sep 23, 2009 #3
    Re: Again an electric potential question ?????

    3. The attempt at a solution

    I know that E in this case = kq/[a*(L+a)]

    a : distance from point P to x2 ( P-x2)
    L : distance from x1 to x2 ( x2-x1)

    The denominator = r²= [a*(L+a)]

    I want Potential V=kq/r

    r=Sqr(The denominator = r²= [a*(L+a)])

    So I just insert the values and got that answer *_* which is wrong
     
  5. Sep 24, 2009 #4
    Re: Again an electric potential question ?????


    V = kq/r is only valid for a point charge or a uniformly charged sphere.
    You have to integrate [itex] E \bullet ds [/itex] along a path from the
    point with x = 5 to infinity.
     
  6. Sep 24, 2009 #5
    Re: Again an electric potential question ?????

    thx 4 clarifying about the potential ^^

    But is'nt taking an integral from 5 to infinity will give me infinity and when i multiply infinity with E i will get also infinity ???


    In case what u r saying is correct what is the value of r then ??

    Is it the distance from point p to x1 or to x2 ? or its something else !

    Appreciate Ur Help in Advance Mr. willem2 :)
     
  7. Sep 24, 2009 #6
    Re: Again an electric potential question ?????

    This computation doesn't have an r in it. you already gave an expression for E without
    R in it. you can integrate the field along the x axis, so E has the same direction as ds, so you
    get

    [tex] \int_5^{\infty} \frac {k q} {(x-x_1)(x-x_2)}dx [/tex]


    This kind of integral with infinity as a limit is called an improper integral and it is actually a
    limit:

    [tex] \int_a^{\infty} f(x) dx = \lim_{L \to +\infty} \int_a^L f(x) dx [/tex]
     
  8. Sep 25, 2009 #7
    Re: Again an electric potential question ?????

    Aha , i get it now. I will submitt the answer & i am sure it will be correct this time , I'll be right back in minutes ^^

    THx again Boss ^^ Really appreciate Ur help !
     
  9. Sep 25, 2009 #8
    Re: Again an electric potential question ?????

    Wow nice Job Friend , its absolutly right ^^

    thank U very much ^^
     
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