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!

Sketch the electical field

  1. Nov 22, 2007 #1
    three small, negatively charged spheres are located at the vertices of an equilateral triangle.The magnitudes of the charges are equal. Sketch the electical field in the region around this charge distribution, including the space inside the triangle.

    ok..so all i know is that density of field lines will be more whr the field is greater..
     
  2. jcsd
  3. Nov 22, 2007 #2
    um..ok..im tryin to work this out..so wht I did i place a positive test charge near each vertex. And since charge flows from th positive to the negative end, the field lines will be drawn from the positve to the negative direction.
     
  4. Nov 22, 2007 #3
  5. Nov 22, 2007 #4
    never mind..i got it..Could someone just check the solutions to these questions...
    1) What is the distance between two protons experiencing a mutually repelling force of magnitude 4.0 x 10-11 N?

    Fe=kq1q2/ r2
    4*10-11= (9*109*1.602*10-19*1.602*10-19)/r2
    r2= (9*1.602*1.602*10-19*10-19*109)/4*10-11
    r2=5.7725*10-18
    r= 2.40*10-9 m

    2) Two point charges, +4.0 x 10-5 C and -1.8 x 10-5 C are placed 24 cm apart. What is the force on a third small charge of magnitude -2.5 x 10-6 C, if it is place on the line joining the other two midway between the originally given pair of charges?

    q1= 4.0*10-3 C
    q2= -1.8*10-5 C
    q3= -2.5*10-6 C

    Fq3q1= Kq3q1/r2
    = 9*109*4*10-5*-2.5*10-6/0.12*0.12
    = -90*10-2/0.12*0.12 = -6250*10-2
    = -62.50 N (left)
    Fq3q2= kq3q2/R2
    = 9*109*-2.5*10-6*-1.8*10-5/0.12*0.12
    = 2812.5*10-2
    = 28.12 N (Left)
    F= Fq3q1+Fq3q2
    = -62.50+28.12= -34.38 N(left)

    3) What are the magnitude and direction of the electric field strength at point Z in the situation below?
    Negatively charged sphere “x” (-2.0 x 10-5 C) is left most, 60 cm to the right of it is a positively charged sphere “y” (8.0 x 10-6C). Point “z” is 30 cm to the right of “y”.

    q1=-2.0*10-5 C
    q2= 8.0*10-6 C

    E1=kq1/r2
    =9*109*-2*10-5/0.90*0.90
    =-18*104/0.90*0.90
    = -222200 N/C (Right)
    E2= kq2/R2
    =9*109*8*10-6/0.30*0.30
    = 72*103/0.30*0.30
    =800000 N/C (Left)
    Hence E on Z= 800000-222200= 577800= -5.8*105 N/C (Left)

    4) If a stationary charged test particle is free to move in an electrical field, in what direction will it begin to travel?
    It would move in the direction of the field.
     
  6. Nov 22, 2007 #5
    just skimming over, the process looks good, but I would watch out for you directions. You won't trip up as easily if you use the MAGNITUDE of the charges when using coulomb's law (just getting the magnitude of the force) and then indicate the direction of the force based on the signs of the charges under consideration.
     
  7. Nov 22, 2007 #6
    i did indicate the directions of the charges..Is question 4 fine?
     
  8. Nov 22, 2007 #7
  9. Nov 23, 2007 #8

    Shooting Star

    User Avatar
    Homework Helper

    2) The forces are in same direction. Should be added.

    3) Take one direction to be the positive one. Here, take it toward the right. E1 is -ve in value, E2 is +ve in value. You have written E2 as +ve to the left, which is incorrect. Final value is correct, since you've written -ve to the left, but a better way would be to fix the +ve dircn beforehand and adjust the sign.

    4) Only when it starts. Afterward, the inertia will be there and the path will not be along the field.
     
  10. Nov 23, 2007 #9
    2) but there is a negative sign. Hence it will result in a subtraction. Wont it?
    3)But if I change the direction, then the direction of the answer will change too. It will become right.
    3) After the field has started, which path will the charge take?
     
  11. Nov 23, 2007 #10

    Shooting Star

    User Avatar
    Homework Helper

    2) The 3rd charge is -ve. The +ve charge will pull the 3rd charge, and the -ve one will push it away. Both forces are in the same dirn.

    3) E2 is acting toward the right.

    4) The motion will depend on the shape of the field.
     
  12. Nov 23, 2007 #11
    2) ok, so the direction of all the forces will be right. But the final F will still involve subtraction, since Fq3q1 is negative. So the answer is right, on the direction has to be right.
    3)so E1 is left, E2 is right, and E on Z is right. (left)

    Also did you check the simulation out?
     
  13. Nov 23, 2007 #12

    Shooting Star

    User Avatar
    Homework Helper

    2) Both the forces are to the left, so you have to add. If you consider right to be +ve, then Fq3q1 -ve and Fq2q1 is also -ve.

    3) If you give proper directions to E1 and E2, the sign of E=E1+E2 will be automatically adjusted. You simply have to sum the two.

    (I'll check out the simulation.)
     
  14. Nov 23, 2007 #13

    Shooting Star

    User Avatar
    Homework Helper

    Yes, I've checked it out. You get a good idea of the electrcic field lines.
     
  15. Nov 23, 2007 #14
    i am so confused...for Q2, if i take as the negative direction to be left. the Fq3q1 will b in the left direction and Fq3q2 will be in the right direction. But if i take into account the charges, then Fq3q1 will be to the right (attraction), and Fq3q2 will also be to the right. So the over all force will be to the right. But although we r adding both the forces, due to the negative sign, there will be a subtraction.
     
  16. Nov 23, 2007 #15
    And for Q3, the direction of E1 is left, E2 is right, and E on Z ir right?
     
  17. Nov 23, 2007 #16

    Doc Al

    User Avatar

    Staff: Mentor

    for Q2

    Assuming the first two charges are placed such that the positive charge is 24 cm to the left of the negative charge, then:
    Fq3q1 will point to the left (it's an attractive force)
    Fq3q2 will also point to the left (it's a repulsive force)

    So both contributions to the net force will be negative.
     
  18. Nov 23, 2007 #17
    So Fq3q1 is -62.50 N (left), Fq3q2 is 28.12 N(left), and hence the net F is -62.50+28.12= -34.38 N (left). Is that right? And what about my posting on the other question..
     
  19. Nov 23, 2007 #18

    Shooting Star

    User Avatar
    Homework Helper

    -62.50-28.12 = -90.62. The minus means to the left.

    (Have patience till tomorrow.)
     
  20. Nov 23, 2007 #19

    Doc Al

    User Avatar

    Staff: Mentor

    No, not right. To the left means negative, so both forces are negative:
    Fq3q1 = -62.50 N
    Fq3q2 = -28.12 N
     
  21. Nov 23, 2007 #20
    ok..so even though we dont get a negative value when we calculate Fq3q2, it is negative because it is acting in the left direction. Is that right? So even in problems invoving charges, we have to take one direction as negative and one as positive..?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Sketch the electical field
  1. Electic Field (Replies: 5)

Loading...