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Point charge on a string in an electric field

  1. Apr 10, 2009 #1
    1. The problem statement, all variables and given/known data
    I need some help solving this question, it is the last one I have to do and I'm not sure how to solve it.

    A point charge (m = 1.0 g) at the end of an insulating string of length 53 cm is observed to be in equilibrium in a uniform horizontal electric field of 9000 N/C with the charge 1.0 cm above the lowest (vertical) position. If the field points to the right determine the magnitude and sign of the point charge.

    Find Magnitude in C:


    2. Relevant equations

    T*cosTheta = mg
    T*sinTheta = F(elec) = qE

    3. The attempt at a solution

    I found this question on a thread from a while ago but I'm not sure entirely what it is saying. Since I have those equations I need to solve for T in the first one to insert it in the second
    T*cosTheta = mg
    T=mg / Cos(theta)

    When I plug that into the second equation
    T*sinTheta = F(elec) = qE

    (mg/cosineTheta) * sineTheta = qE = F(elec)

    My question is, do I need all three parts of this last equation? Or can I simplify it to this:
    (mg/cosineTheta) * sineTheta = qE

    Even if I do this, how do I figure out the angle?
    And in this case what does E represent?
     
  2. jcsd
  3. Apr 10, 2009 #2
    I tried this, but the answer was incorrect. Can you see what I did wrong?


    | \
    | \
    | \
    52 | \ 53
    | \
    | T \
    *

    So the point charge is basically the dobber of a simple pendulum. In this case you have two forces acting on it: gravity, and electrostatic force. Electrostatic force can be described as:

    F = E * q

    where F is the force vector, E is the electric field vector and q is the charge on the particle. This force vector is directed parallel with the electric field (to the right).

    So now look at gravity:

    F = m * g

    where m = mass of the particle and g = force of gravity (9.8 kg/s^2). Give a 1 g (.001kg) particle, the gravitational force is .0098 N). This is directed straight down (toward the earth).

    The string will counter-act both of these forces and can be described as:

    Fx = F cos T
    Fy = F sin T

    Give the diagram above:

    sin T = 52 / 53
    (because the particle is 1 cm above the lowest point and the string is 53cm long)

    So, Fy has to counteract gravity:

    Fy = F sin T = m * g
    F * (52/ 53) = 0.0098 N
    F = 0.0098 * 53 / 52

    So Fx has to counteract the electrostatic force:

    Fx = F cos T = E * q

    Solve for q as:

    q = (F cos T) / E

    Where F = 0.0098 * 53 / 52
    T = arcsin (52/53)
    E = 9000 N/C

    Plugging in these values:

    1.376 * [cosine{arcsin (52/53)}]
    1.376 * .999999
    q=1.375999/9000
    q=1.5288e-4

    But this is wrong for q, does anyone one know what went wrong?
     
  4. Apr 10, 2009 #3

    LowlyPion

    User Avatar
    Homework Helper

    The first thing to do is figure the angle.

    That's determined by the height of 1 cm.

    That means that the Cos of the angle with the vertical must be

    (53-1)/53 = 52/53

    Cos-1(52/53) = angle

    The Tangent of that angle is the ratio of the 2 forces ... mg down and the Fe horizontally - then isn't it?
     
  5. Apr 10, 2009 #4
    Ok so cos-1 (52/53) = 11.14 degrees

    So can I plug that into the equations I was given up above?

    T*cos(11.14) = mg
    T*sin(11.14) = F(elec) = qE
     
  6. Apr 10, 2009 #5
    I would need to convert g=9.81 m/s to g=0.00981 km/s to get coulombs correct?
    If that does work,
    T= (0.00981 * 1)/cos(11.14)
    T=.009998389

    T*sin(11.14) = F(elec) = qE
    .009998389 * sin(11.14) =qE
    [.009998389 * sin(11.14)]/E = q
    .0019317587 / E = q
    .0019317587 / 9000 = q
    2.146398 e-7

    Correct?
     
  7. Apr 10, 2009 #6

    LowlyPion

    User Avatar
    Homework Helper

    I'd go from here with T = mg/cos(11.14)
    So ...

    q*E = tan(11.14)*mg
     
  8. Apr 10, 2009 #7

    LowlyPion

    User Avatar
    Homework Helper

    I'd use the previous equation in SI units and then the answer is already in C.

    q = .197*.001*9.81/9000

    It yields your answer.

    But I would express it as .215 μC

    The answer they are looking for is in C though.
     
    Last edited: Apr 10, 2009
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