1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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

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


    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*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

  7. Apr 10, 2009 #6


    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


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook