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Electrostatics questions, parallel plates, force, field

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data

    1.
    A 2*10^-6C charged object is .40m left of an another 2*10^-6C charged object, with a 3*10^-6C charged object .40m to the right of the centre object, creating three charged objects in a line. Calculate the magnitude of the net electric force on the centre charge due to the other two charges.

    2*10^-6C ------.40m-------2*10^-6C------.40m----3*10^-6C

    2. An electron travelling horizontally at a speed of 8.70 * 10^6m/s enters into an electric field of 1.32 *10^3 n/c between two horizontal, parallel plates. The plates are 14.0cm long and are seperated by a distance of 8.00cm. Calculate the vertical displacement of the electron through the plates. [3*10^-2m]

    3. A 3uC (A) charge is .333m above charged B (3uC), with charge C (3uC) 0.333m left to charge b, creating a triangle. Find the net electric force on charge C.


    4. Two point charges are initially 6.0cm apart, and are then moved so that they are 2.0cm apart. If the initial force between these two point charges was F, what is the new force [9F]

    [number inside these are the answers]


    2. Relevant equations
    E=fe/q
    E=kq1q2/r^2
    E=kq/r^2
    V=E(energy)/q
    E=v/d
    k=8.99*10^9
    charge of an electron= 1.6*10^-19
    mass of an electron = 9.11*10^-31kg


    3. The attempt at a solution

    1.
    E1= (8.99^10^9)(2*10^-6)(3*10^-6)/.40^2 = .337125
    E2= (8.99*10^9)(2*10^-6)(2*10^-6)/.40^2= .22475
    E1+E2=.561875N

    I added them together based on coulombs law, charged A and C are repelled and the centre charge is also repelled. I dont have the answer so i dont know if im right

    2. First i found acceleration

    a=Fe/m=Eq/m=(1.32*10^3 * 1.6*10^-19)/9.11*10^-31 = 2.318331504*10^14

    I actually dont know where to go from here

    using projectile motion i listed the x and y variables

    x:
    Vi= 8.7*10^6
    d=.14m

    y:
    a=2.31...*10^14
    d=.08m


    how do i find time? then the displacement? im a litte confused any help please


    3.
    F(BonC) = (8.99*10^9)(3*10^-6)(3*10^-6)/.333^2 = .7296485675

    then for AonC

    First i use pytha to find the length of the hypoteneuse connecting A and C.

    [itex]\sqrt{.333^2 + .333^2}[/itex] =.4709331163

    then used tan to find the angle, .333/.333= 1 tan-1(1) = 45, which makes the opposite triangle 45 degrees as well.

    Then found the force on the hyp.

    F= (8.99*10^9)(3*10^-6)(3*10^-6)/.4709..^2 = .3648242837

    Now based on the diagram i have, , the x and y value triangle will use sin for x and cos for y based on opposite and adjacent, i just need some confirmation on this.
    Also when i found the x and y values, would i subtract or add the x value of the force (BonC)?

    A
    |
    |
    |
    B---------C
     
  2. jcsd
  3. Oct 15, 2012 #2
    How long does it take the electron to travel along the plates?
     
  4. Oct 16, 2012 #3
    it doesnt tell you you have to find it, and i do not know how well its usually t=[itex]\sqrt{2d/a}[/itex] however i can never come up with the right answer

    based off d=vit+1/2at^2
     
  5. Oct 16, 2012 #4
    The electron is travelling horizontally at a given speed along horizontal plates of a given length. Can't be too hard.
     
  6. Oct 16, 2012 #5
    yes but i need to find the vertical displacement using projectile motion techniques try it
     
  7. Oct 16, 2012 #6
    You said you cannot find time. What I said in #4 is enough to find time.
     
  8. Oct 16, 2012 #7
    Well i cant get it so try helping?
     
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