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Physics problem

  1. Dec 20, 2003 #1
    Problem 16.
    Three charges: +8.2uC 4.8 cm to the left of 4uC and a -2.3uC 2.1 cm to the right of the 4uC charge.
    What is the electric field strength at a point 2.7cm to the left of the middle charge? In N/C.
    Note" Is the answer 1.733218664*10^7 from the 8.99*10^9 (4*10^-6)/(0.027^2)=49327846.36 and (8.99*10^9)(8.2*10^-6)/(0.048^2)=31995659.72 which were subtracted to get that answer.
  2. jcsd
  3. Dec 20, 2003 #2
    You might want to look that over again.

    For starters, there are 3 charges & you figured the field from only 2 (& got one of the distances wrong).

    Draw a diagram first, & work from that.
  4. Dec 20, 2003 #3
    I drew a diagram and have 8.2 uC 2.1 cm away from the point that is 2.7 cm to the left of the middle charge. In addition should i multiply 8.99*10^9 to -2.3uC divided by 3.1 cm and subtract what is now three values to get my answer?
  5. Dec 20, 2003 #4
    First, where did you get 3.1 cm?

    Second, make sure you keep the directions straight. The field from the positive 8.2 μC charge is directed toward the right. What are the directions of the other two fields?
    i.e.: same direction = add; opposite direction = subtract

    And don't forget, you are dividing by the square of the distance.
  6. Dec 20, 2003 #5
    woops! From -2.3uC to 4 uC is 2.1cm and from 4 uC it is 2.7 cm to reach that point so the distance would be 4.8cm. So for -2.3uC I multiply 8.99*10^9 to -2.3uC divided by 4.8 cm or 0.048m?
  7. Dec 20, 2003 #6

    But think carefully about the directions. It's not just a question of the sign of the charge. You have to consider the relative positions of the particles.
  8. Dec 20, 2003 #7
    The field from the +8.2 μC charge has the same direction at point P as the -2.3 μC charge.

    Do you see why?
  9. Dec 20, 2003 #8
    No, i don't see why 8.2uC and -2.3uC have the same direction. I got three values from 4*106-6 is 49327846.36, from 8.2*10^-6 is 167160997.7, and -2.3*10^-6 is 8974392.361. So would i add 167160997.7 to 8974392.361 and subtract that from 167160997.7/
  10. Dec 20, 2003 #9

    The direction of the electric field at any point P is the same as the direction of the electrical force that would be experienced by a positive "test" charge placed at that point.

    In this problem, if you placed a positive test charge at point P, it would be repelled by the positive 8.2 μC charge (call that charge A) towards the right since they're both positive, so the component of the field from charge A at point P is directed toward the right.

    But the negative 2.3 μC charge (call it C) would ATTRACT a positive charge, so a positive test charge located at point P would be pulled to the right. Therefore, the field component produced by charge C at point P is also directed toward the right.

    Therefore, the field components of charges A and C at point P are added, not subtracted.

    On the other hand, what would charge B (the +4 μC charge in the middle) do to a positive test charge at point P?

    Get it?
  11. Dec 20, 2003 #10
    Are the three values I got from the charges right?
  12. Dec 20, 2003 #11
    Yes. Now you just have to figure out what to add & what to subtract.
  13. Dec 20, 2003 #12
    So gnome since you said that "The field from the +8.2 ìC charge has the same direction at point P as the -2.3 ìC charge." I would add
    167160997.7 to 8974392.361 to get 176135390.1. From that I would subtract 49327846.36 and get 126807543.7, right?
  14. Dec 20, 2003 #13
    Yes, but do you understand why, or are you just taking my word for it?
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