1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

CAN SUMBODY HELP EM WID DIS PROBLEMss ZZZZZZz life or death situation!

  1. Jan 10, 2005 #1
    1. The electrostatic force between two point charges is 1.2 x 10^-2 N. If the distance between them is doubled, the charge of one of the point’s doubles and the charge of the other point triples. What will be the force between them?
    Kq1q2/r2

    3. Two charges one charge +1.5 x 10^-2 C and the other charge -2.7 x 10^-5 C, are 20.0 cm apart. The positive charge is to the left of the negative charge.
    a) Draw a diagram showing the point charges and label a point Y that is 5.0 cm away from the positive charge, on the line connect the charge (field lines need not be drawn.)
    b) b.) Calculate the electric field at point Y


    4. Find the final speed of an electron, starting form rest, passing between two parallel plates with a potential difference of 4.5 x 10^3 V. An electron mass of 9.1 x 10^-31 kg and a charge of 1.6 x 10^-19 C. Is this answer valid? Explain.

    (yes it is valid because …)

    5. Two parallel plates labeled W and X are separated by 5.2 cm. The electric potential between the plates is 150V. An electron starts from rest at time tW and reaches plate X at time tX. The electron continues through the opening and reaches point P at time tp (remember e = - 1.6 X 10^-19 C and the mass of an electron is 9.1 X 10^-31 kg)

    a.) Sketch the speed-time graph on the axes below
    b.) Determine the kinetic energy of the electron as it arrives at plate X
     
  2. jcsd
  3. Jan 10, 2005 #2
    show your work. read the sticky...
     
  4. Jan 10, 2005 #3
    what sticky?
     
  5. Jan 10, 2005 #4
    which part do you NOT understand? it seems like you have the formulas for problem 1 already.... and the rest seems like a graphing problem, just follow the question, draw the graph, and apply the formulas in your text books, that's it, do you not understand how to graph or what? Let me know what you do not understand so that i can help you.....I don't wanna just hand out the answer
     
  6. Jan 10, 2005 #5
  7. Jan 10, 2005 #6
    For question 1:
    K doesn't change, q1 is doubled so it's 2, q2 is tripled, so it's now 3 and r doubled, so the new force is 2*3*k/4 which is 3/2 of the old one. So just multiply the 1.2 x 10^-2 N by 3/2. And the answer wud be 1.8x10^-2N


    FOR QUESTION 3
    For 3. I have no idea how to do the diagram I am like really lost L, for the electric field at point Y I have E on point y = kq1/r^2 = (9.0 x 10^9 Nm^2/c^2) (1.5 X10^-2)/ 0.05^2 and I got 5.4 X10^10N but I feel like something is missing.
     
  8. Jan 10, 2005 #7
    sorry guys i am new..and i am not trying to get somebody to do my homework i wanna understand it myself i am just a little slow i would really appreciate it if you guys could help me, i will post my answers thus far for 4 and 5 next
     
  9. Jan 10, 2005 #8
    FOR QUESTION 4


    Q deltaV /r = ½ kmv^2

    (1.6 X 10^ -19 C ) (4.5 X 10^3V) / 3,2 X10 ^-13 = ½ x 9.1 x 10^-31 X v^2
    v^2 = 4.94 X 10^24
    v = &.03 X 10^13


    yes this is possible…but why?
     
  10. Jan 10, 2005 #9
    FOR QUESTION 5

    FE = Ek
    Q delta V/r = 1/3 mv^2

    1.6 x 10^-19 C (150V) / 0.05a = ½ (9.1X10^-31 kg) (v^2)
    V= 2.9 X 10^-23 m/s
     
  11. Jan 10, 2005 #10
    i am not going to check your arithmatic...coz i don't have a calculater... sorry about that.. the first problem you did is totally correct, the second one is wrong, how? the positive charge and negative charge are separate 20cm, point Y is located between two charges and 5cm away from the negative one, how far away is it from the positive one? the E field @ Y is just the sum of the E field of two charges.... the last one, assume your v is right(i am not checking your arithimatic) , how is it compare with speed of light c= 3*10^8 m/s, notice nothing can travel faster than light
     
  12. Jan 10, 2005 #11
    Q delta V/r = 1/3 mv^2
    should be one half @right hand side instead of one 3rd, but i think it is a typo...

    1.6 x 10^-19 C (150V) / 0.05a = ½ (9.1X10^-31 kg) (v^2)
    this is correct...if you change the a to 2......

    V= 2.9 X 10^-23 m/s
    WRONG, obviously the v is too small.... some arithmatic mistake, do it carefully
    and, isn't that the problem asking you the kinetics energy... why are you doing speed here?
     
  13. Jan 10, 2005 #12
    oh thanx buddy ur awesome!
     
  14. Jan 10, 2005 #13
    oh, sorry, one more mistake you have made..
    because i didn't check your number, i missed this one.....

    Q deltaV /r = ½ kmv^2===>>wrong
    the right one should be
    qV=1/2mv^2
    where did you get 3.2*10^-13 in
    (1.6 X 10^ -19 C ) (4.5 X 10^3V) / 3,2 X10 ^-13 = ½ x 9.1 x 10^-31 X v^2
     
  15. Jan 10, 2005 #14
    4. Find the final speed of an electron, starting form rest, passing between two parallel plates with a potential difference of 4.5 x 10^3 V. An electron mass of 9.1 x 10^-31 kg and a charge of 1.6 x 10^-19 C. Is this answer valid? Explain.

    wut is the value fo r in this question?
     
  16. Jan 10, 2005 #15
    you dont even need r
    energy is qV, no matter what r is, the energy will be the same
     
  17. Jan 10, 2005 #16
    i dunt understand
     
  18. Jan 10, 2005 #17
    but for question 3 i got - 5.39903 x 10^14 n/c
    does dat seem right?
     
  19. Jan 10, 2005 #18
    oh man am so confused
     
  20. Jan 10, 2005 #19
    your formulas is wrong
    you should use
    qV=1/2mv^2 instead of qV/r=1/2mv^2
    since the potential energy is independent of the distance r, It related only to q and V...
     
  21. Jan 10, 2005 #20
    ohhhhhhhhh i c
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?