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Magnitude of electric field

  1. Aug 16, 2010 #1
    1. The problem statement, all variables and given/known data

    If Q = 80 nC, a = 3.0 m, and b = 4.0 m in the figure, what is the magnitude of the electric field at point P?

    [PLAIN]http://img829.imageshack.us/img829/9703/72660756.gif [Broken]

    2. Relevant equations

    [tex]\vec{E}=k_e \frac{q}{r^2}\hat{r}[/tex]

    3. The attempt at a solution

    The magnitude of the electric field at P due to the first charge is:

    [tex]E_1 = (8.9 \times 10^9) \frac{(2 \times 80) \times 10^{-9}}{(4^2+3^2)}=56.96[/tex]

    due to the second charge it's:

    [tex]E_2 = (8.9 \times 10^9) \frac{(-80) \times 10^{-9}}{(4^2)} = - 44.5[/tex]

    And due to charge 3:

    [tex]E_3 = (8.9 \times 10^9) \frac{(2 \times 80) \times 10^{-9}}{(4^2+3^2)}=56.96[/tex]

    The total magnitude of the electric field at P would be: [tex]56.96-44.5+56.96 = 69.3[/tex]

    But the correct answer is 47 N/C. Could anyone please show me what is wrong with my calculations?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 16, 2010 #2
    You must treat electric field as a vector! Draw the electric field vectors of each charge and with a little bit of trigonometry it should be no problem.
     
  4. Aug 16, 2010 #3

    collinsmark

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    Hello roam,

    You need to treat the individual electric field contributions as vectors. Note the unit vector [itex] \hat r [/itex] in your

    [tex]
    \vec{E}=k_e \frac{q}{r^2}\hat{r}
    [/tex]

    equation. This unit vector is different for each charge you consider. It has both x and y components.

    Express each electric field vector in terms of its x and y components, and then add them together. :wink:
     
  5. Aug 16, 2010 #4
    Hi Collinsmark

    I tried that, it didn't work:

    [tex]\vec{E_1}=56.96 cos (90) \hat{i}+ 56.96 sin (90) \hat{j}[/tex]

    [tex]= 56.96 \hat{j}[/tex]

    [tex]\vec{E_2}=-44.5 cos (0) \hat{i}-44.5 sin (0) \hat{j})[/tex]

    [tex]= -44.5 \hat{i}[/tex]

    [tex]\vec{E_3}=56.96 cos (90) \hat{i}+ 56.96 sin (90) \hat{j}[/tex]

    [tex]= 56.96 \hat{j}[/tex]

    And now adding up to find the components of the net electric field vector::

    [tex]E_x= -44.5[/tex]

    [tex]E_y= 56.96+56.96=113.92[/tex]

    The resultant would be 122.30 which is wrong, the right answer is the single number 47 N/C. So why am I not getting the right answer?
     
  6. Aug 16, 2010 #5

    collinsmark

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    The direction of [itex] \hat r [/itex] is from the charge, to the test point. So the [itex] \hat r [/itex] direction of the top charge is [itex] \hat r = 4/5 \ \hat \imath - 3/5 \ \hat \jmath [/itex]. I'll let you do the rest.
     
  7. Aug 16, 2010 #6
    So [tex]\hat{r}[/tex] is a unit vector directed from the charge toward P. But I'm a little bit confused... where did you get "4/5" and "-3/5" from? I know that 5 is the hypotenuse (line from the top charge to P), but why did you divide them like that? :confused:
     
  8. Aug 16, 2010 #7

    collinsmark

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    Well, putting it somewhat simply, 'sine' is the opposite over hypotenuse. 'Cosine' is the adjacent over hypotenuse. There is a negative sign associated with the [itex] \hat \jmath [/itex] since that component is pointing down.

    [tex] \sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} [/tex]

    [tex] \cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} [/tex]
     
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