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Magnitude and Direction of Electric Fields

  1. Jan 11, 2008 #1
    1. The problem statement, all variables and given/known data

    A small object A, electrically charged, creates an electric field. At a point P located 0.250m directly north of A, the field has a value of 40.0N/C directed to the south.

    A) 1.11×10^−9C
    B) −1.11×10^−9C
    C) 2.78×10^−10C
    D) −2.78×10^−10C
    E) 5.75×10^12C
    F) −5.75×10^12C


    What is the charge of object A?

    2. Relevant equations

    E = (KQ)/r^2

    3. The attempt at a solution

    After manipulating the above equation, I found out that the charge Q = (E*r^2)/K

    Q = (E*r^2)/K
    = (40*(0.25^2))/9×10^9
    = 2.78×10^−10C

    makes sense to me, but should it be negative or positive?
     
  2. jcsd
  3. Jan 11, 2008 #2

    dynamicsolo

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    Remember that the vector direction of electric field is defined as radially outward from a positive charge and radially inward from a negative charge. The point P is north of the charge and the field vector there points back southward toward the charge. So what sign would the charge need to have? (BTW, I agree with your charge magnitude.)
     
    Last edited: Jan 11, 2008
  4. Jan 11, 2008 #3
    A way to remember the rule that dynamicsolo stated is to recall that E fields are defined in terms of the force felt by a unit test charge, which is positive by assumption. That gives the directions stated above.
     
  5. Jan 11, 2008 #4

    dynamicsolo

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    Thanks, I did omit to say that, since it is properly part of that definition.
     
  6. Jan 11, 2008 #5
    so it would be positive?
     
  7. Jan 12, 2008 #6

    Dick

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    Why would you think that?
     
  8. Jan 12, 2008 #7
    cse63146 - Think it through in these steps:

    - Is the test charge (i.e. field point in question) North or South of the field source at A?
    - Given that location, does the direction of the force it feels indicate an attractive or repulsive force?
    - Considering that the test charge is defined to be positive, what does your previous answer tell you about the sign of the source charge at A?
     
  9. Jan 12, 2008 #8
    Since there's a point P north of charge A, and the electric field is directed to the south (towards the charge), it looks like it's inward, which makes the charge negative?
     
  10. Jan 12, 2008 #9

    dynamicsolo

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    That would be correct.
     
  11. Jan 12, 2008 #10
    There's a part B of the question which only showed up after you answer part A.

    If a second object B with the same charge as A is placed at 0.250m south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P?

    I know that E = (KQ)/r^2

    so

    E = (KQ)/r^2
    E = ((9×10^9)(2.78×10^−10)/0.5^2
    E = 10

    So the total magnitude produced by the two objects 10 + 40 = 50N/C?
     
  12. Jan 12, 2008 #11
    Correct.
     
  13. Jan 12, 2008 #12
    Got both questions right. Thank you all.
     
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