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Electric Potential

  1. Feb 26, 2008 #1

    tony873004

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    If [tex]\overrightarrow E = \left( {16\,{\rm{V/m}}} \right)\,{\rm{\hat i + }}\left( {{\rm{8}}{\rm{.5 V/m}}} \right){\rm{\hat j}}[/tex] and the potential is 0 at the origin, find the potential at point P with coordinates x=1.5 m, y=3.5 m.

    I made a diagram. (each tick mark = 2 units). I imagine this is just a uniform electric field. The equipotential lines are perpendicular to the field lines. The two lines going up and to the right are field lines, and the two perpendicular lines are lines of equipotential. The one passing through the origin is 0V. The other one is passing through the point (1.5, 3.5), but I don't know its value. Don't I need to know how far away the charge is that is creating the Electric field in order to determine how potential will change with distance?
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    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 27, 2008 #2
    You can set this up as a line integral (if you're doing calculus based physics). That is, pick your path ([itex]\vec{r}(t)=(1.5t,3.5t,0)[/itex] with [itex]0\le t\le1[/itex] would be a good one) and use

    [tex]V_f-V_i=-\int \vec{E}\circ d\vec{r}[/tex].
     
  4. Feb 27, 2008 #3

    tony873004

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    I don't understand that.

    I'm still stuck on this. All I can think to do is compute the magnitude of the electric field: sqrt(16^2 + 8.5^2), and the distance from point P to the origin: sqrt(1.5^2 + 3.5^2), and since V=Ed, multiply them together which gives me 70V. But the answer in the back of the book is -54V.
     
  5. Feb 27, 2008 #4

    tony873004

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    I got it! It's dot product of E*V, not regular old multiplication of |E|*|V| which is what I was doing. Now I get -54 like the back of the book.
     
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