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Find strength and direction of electric field

  1. Sep 4, 2010 #1
    1. The problem statement, all variables and given/known data
    There is a charge of -4.05mC at (3,0)m and a charge of +3.85mC at (10,0)m. And I have to find the strength and direction of the field at (2.5,7)m. (Direction to be in degrees).

    2. Relevant equations


    3. The attempt at a solution

    I've found the fields to be 7.40x10^5 N/C for the first and 3.28x10^5 N/C for the second on the point.

    And I'm stuck from here. I've started figuring out all sorts of angles but I just get swamped. Could someone please nudge me in the right direction?
  2. jcsd
  3. Sep 4, 2010 #2


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    Homework Helper

    Can you show your calculations of the fields?
  4. Sep 4, 2010 #3
    Sure thing, for the first:
    (9x109)(4.05x10-3 / 49.25

    The 49.25 came from the distance (2.5-3)2+(7-0)2 (the sqrt of this and then squaring again would cancel each other so I didn't bother with it)

    So, that ended up being (3.645x107)/49.25 equaling 740,102=7.40x105

    For the second:

    The 105.25 coming from (2.5-10)2+(7-0)2) again ignoring the sqrt squared and getting 56.25+49=105.25

    And so, (3.465x107)/105.25 = 328,361 = 3.28x105
  5. Sep 4, 2010 #4


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    OK. Now the electric field E1 due to the first charge is towards the charge and the field E2 due to the second charge is away from the charge. To find the resultant we must know the angels made by these fields with horizontal.
    For the Ε1, tanθ1 = 7/0.5 and for E2, tanθ2 = 7/7.5. Find θ1 and θ2.

    Now take the horizontal and vertical components of E1 and E2 and find the resultant E.
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