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Strength of the electric field

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data

    a) What is the strength of the electric field at the position indicated by the dot in the figure?
    b) What is the direction of the electric field at the position indicated by the dot in the figure? Specify the direction as an angle above the horizontal line.
    Figure attached

    2. Relevant equations

    E = Kq/r^2


    3. The attempt at a solution

    At first I thought strength of the electric field on both side would be same and the net strength would be O
    then I calculated the distance from the point to the charge using Pythagoras theorem and then using the above equation which give me E as 1836 N/C. so for both side it would be 1836*2 = 3673 N/C

    but the answer was wrong?
    where did I mess up?
     
  2. jcsd
  3. Mar 15, 2009 #2

    LowlyPion

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    I'm wondering the same thing.
     
  4. Mar 15, 2009 #3
    ya i tried both 0 and 3.6*10^3
    but didnt work
     
  5. Mar 15, 2009 #4

    LowlyPion

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    Well let's see here.

    There's no figure.
    There's no calculations.
    There's no real description of what you are trying to calculate.

    Can you imagine the difficulties others might have in trying to help you yet?
     
  6. Mar 15, 2009 #5
    sorry i thought i attached the figure.

    using Pythagoras theorem i figured out the distance from the point to the charge which is 7.07 cm = 0.07 m

    and then i used the equation,
    E = kq/r^2 = (9*10^9 Nm^2/C^2)(1*10^9 C)/(0.07 m)^2 = 1836.7 N/C
    so for both side it would be 2*1836.7 N/C

    sorry for the inconvenience.
     

    Attached Files:

  7. Mar 15, 2009 #6

    LowlyPion

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    I can't see the figure yet. So do you have multiple charges? Can you describe it a little more?
     
  8. Mar 15, 2009 #7
    ya i have two charges with with +1 nC charge which is 10 cm away from each other. there is a poitn in the middle 5 cm away. its like a triangle. they ask me to measure the strength and direction of the electric field.

    Kinda look like this. I have attached the pic in last reply

    O +1 nC
    .
    .
    .
    5 cm
    .
    .
    .
    ------5cm----- o
    .
    .
    .
    5 cm
    .
    .
    .
    O +1 nC
     
  9. Mar 15, 2009 #8

    LowlyPion

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    Well there's the problem.

    Remember the E field is a vector field. When using superposition you can't just add the |E| of both vectors.

    Add x,y components only.
     
  10. Mar 15, 2009 #9
    how do we figure out x component?
     
  11. Mar 15, 2009 #10

    LowlyPion

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    You have the angle for both E's
     
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