Magnitude and Direction of Electric Fields

  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. dynamicsolo

    dynamicsolo 1,656
    Homework Helper

    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. 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. dynamicsolo

    dynamicsolo 1,656
    Homework Helper

    Thanks, I did omit to say that, since it is properly part of that definition.
  6. so it would be positive?
  7. Dick

    Dick 25,855
    Science Advisor
    Homework Helper

    Why would you think that?
  8. 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. 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. dynamicsolo

    dynamicsolo 1,656
    Homework Helper

    That would be correct.
  11. 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


    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. Correct.
  13. Got both questions right. Thank you all.
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?