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Electric Field at the center of a square

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data

    Calculate the electric field (magnitude & direction) at the center of a square 52.5 cm on a side if one corner is occupied by a = -38.6 µC charge and the other three are occupied by -27.8 µC charges.


    2. Relevant equations
    Enet= E1center + E2center + E3center + E4center
    E2,3,4= k |Q| / (length*sqrt(2)/2)2
    E1=k |Q| / (length*sqrt(2)/2)2

    3. The attempt at a solution

    E1=k |Q| / (length*sqrt(2)/2)2
    E1=(8.99E9 Nm2/C2)(38.6E-6 C) / (0.525*sqrt(2)/2)2=2.51E6 N/C

    E2,3,4= k |Q| / (length*sqrt(2)/2)2
    E2,3,4=(8.99E9 Nm2/C2)(27.9E-6 C) / (0.525*sqrt(2)/2)2=1.82E6 N/C

    Enet= 7.97E6 N/C

    But according to Mastering Physics, the answer is 6.98E5 N/C. What did I do wrong?
    And I'm thinking that direction of electric field is towards the -38.6 µC charge. Is that correct?
     
  2. jcsd
  3. Sep 4, 2011 #2

    tiny-tim

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    welcome to pf!

    hi format1998! welcome to pf! :smile:
    i don't undestand where 7.97 comes from :redface:

    but anyway, electric fields are vectors, and you have to add them as vectors

    try again :smile:
     
  4. Sep 4, 2011 #3
    So was I making a mistake in calculating the distance from the corner of the square to the center. Is the distance suppose to be...

    d = (.525m/2)/cos(45degrees)

    I calculated...
    E1=(8.99E9 Nm2/C)(38.6E-6 C)/(.3712)= 2.52E6 N/C
    E2,3,4=(8.99E9 Nm2/C)(27.9E-6 C)/(.3712)= 1.82E6 N/C

    Then I added all Enet= E1 + 3*E2,3,4 and I got 7.98E6 N/C

    But Mastering Physics says the correct answer is 6.98E5 N/C. What am I doing wrong? Any and all help is greatly appreciated. Thank you very much in advance!
     
  5. Sep 5, 2011 #4

    tiny-tim

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    hiformat1998! :smile:
    you've added them as scalars!!

    electric potentials are scalars, so that would be ok

    but electric fields are vectors (like forces), and you have to add them as vectors
     
  6. Sep 5, 2011 #5
    I feel so dumb now but I don't know what to do when you sy add them as vectors. I thought since I calculated all the individual Electric Field's as vectors that should be enough. My professor did em that way. Please tell me what I need to do...

    Any and all help is appreciated. Thank you so much in advance.
     
  7. Sep 5, 2011 #6

    tiny-tim

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    hi format1998! :smile:
    a vector has coordinates

    you have to specify the coordinates if you want to add them

    (the magnitudes are not enough)

    so write each of the three force as an (x,y) pair of coordinates, then add those three pairs :wink:
     
  8. Sep 6, 2011 #7
    This is quite similar to your 'electric force acting on electron' question, and is easiest answered using the vector notation I used in my reply (I'll let you off not using your lecturer's way for now!).

    The separation between each charge and the centre is always the same, as you've shown. You've now got to pay some attention to working out these unit vectors. Remember that the vector always points from the charge towards whatever you're interested in. So for your -38.6uC charge, the vector will be 26.25x - 26.25y. I'll let you work out the others and see what you get!
     
  9. Sep 6, 2011 #8
    So I understand that the component of the vector for the -38.6uC charge is as follows:

    x-component = 0.2625m and the y-component = 0.2625m and the angle is 45 degrees. I still don't understand what to do with it.
    I calculate E= k (|Q|/r^2) for the magnitude ... in order to get the actual vector what am I suppose to do w/ it? Does my question make sense?
     
  10. Sep 6, 2011 #9
    You'll probably find it easier if you use the 'unit vector' notation, rather than your lecturer's 'degree' notation for this one.

    So you've got your vector for the first charge (=26.25x - 26.25y), and calculate the unit vector by dividing 26.25x - 26.25y by its magnitude. You then multiply the magnitude of the electric field by this unit vector, and this tells you what the electric field is at the centre of the square, caused by this charge, and which direction it is pointing. If you repeat this process for the three other charges, you can simply add the 4 fields together, remembering that you can only add x's to x's and y's to y's.

    Keep asking questions if you don't understand!
     
  11. Sep 6, 2011 #10

    Sorry to sound so dumb but please clarify the above quote. Did you mean ||v||=sqrt((.2625)^2+(-.2625)^2) = 0.371 I don't understand how I'm suppose to divide through... do you mean take the x component and divide by the unit vector .371 and do the same to the y-component?

    How do I get the number that I need to multiply with the magnitude of the Electric Field?
     
  12. Sep 6, 2011 #11
    Just think of 26.25x - 26.25y as an equation with x and y as the variables. I'll work in metres so as not to confuse you, so the vector becomes 0.2625x - 0.2625y. As you correctly said, the magnitude is 0.371, so dividing the vector by 0.371 gives you [itex]\frac{0.2625x - 0.2625y}{0.371}= 0.708x - 0.708y[/itex]. By multiplying the magnitude of the E field by this, you will end up with two terms; an x one and a y one.

    What you need to do now is find the magnitudes of the electric fields for each of the other charges, and also calculate their unit vectors. Multiply each E field magnitude by the appropriate unit vector, and you will see that you have an x and y component for each E field. Then add these four equations together. Again, by treating x and y as variables in an equation, you can do a bit of factorisation to get two terms: Ax + By = Enet.

    Keep asking if you're not sure!
     
  13. Sep 6, 2011 #12
    See attached file for what I did ..... I'm still doing something wrong because I'm still not getting 6.98E5 N/C (which is the answer given by mastering physics)
     

    Attached Files:

  14. Sep 7, 2011 #13

    tiny-tim

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    hi format1998! :smile:

    i haven't checked your calculations,

    but since it's obvious what direction the total field is, wouldn't it be a lot easier just to use components in that direction (rather than along the sides)? :wink:
     
  15. Sep 7, 2011 #14
    To give you an even bigger hint, it is much easier to turn the square onto its side so it looks like a diamond. By doing this you will only have to calculate the magnitude of the electric field for two of the charges (and you don't have to use angles or vectors at all!).
     
  16. Sep 9, 2011 #15
    I figured it all out, finally. Thank you so much for all of your help.
     
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