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Electric fields

  1. Oct 28, 2004 #1
    I'm doing the following problem, it seems to me that I am doing all the correct steps but my answer differs from the answer key, am I missing something?

    1. What is the magnitude and the direction of the electric field at a point midway in between a -8.0uC and a +6.0uC change 4cm apart.


    This is what I'm doing:

    E = ( k * (Q1 / r1^2) ) * ( 1 + ( (Q2 / Q1) / (r2^2 / r1^2) ) )

    ==

    (9x10^9 * ( (8 * 10^-6) / ( 2 * 10^-2) ) ) * ( 1 + ( (6 / 8) / (2 / 2) ) )

    ==

    3600000 * 1.75 = 6300000 N/C


    now..... that is what I'm getting, could somoen please point out to me what I am forgetting to do, if anything.



    __________edit

    stupidity strikes again..... forgot to square r1 :yuck:
     
    Last edited: Oct 28, 2004
  2. jcsd
  3. Oct 29, 2004 #2

    Galileo

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    Why write E in such a strange way? (to me anyway)

    Use Coulomb's law and the superposition principle.
    Calculate the field due to Q1 and the field due to Q2, then add.
     
  4. Oct 29, 2004 #3
    Hello joej...

    The formula for electric field at a point distant r units from a charge Q is:

    [tex]\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^2}\hat{e}_{r}[/tex]

    The direction of the field is radial. In your problem you need to compute

    [tex]\vec{E}_{net} = \vec{E_{1}} + \vec{E_{2}}[/tex]

    the two terms arising due to the electric fields produced (independently--make a note of this, the superposition principle) has been used here) by the two charges. Have you used this fact? :confused:

    Cheers
    Vivek
     
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