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!

Particle around pivot in uniform E

  1. Sep 25, 2009 #1
    Decided to repost for general help with this problem.

    1. The problem statement, all variables and given/known data
    A particle with charge q and mass m is tied by a (I assumed stiff) string of length L to a pivot point P, all of which lie on a horizontal plane. A uniform electric field E is placed over this system. If the initial position of the particle is at a point where the string is displaced T degrees from the axis parallel to the electric field, what will be the speed of the particle when it reaches this axis? The particle is initially at rest. Assume no outside work is done on the particle.

    E = 250 V/m
    q = 1.8 micro C
    L = 1.3 m
    m = .015 kg
    T = 50 degrees
    v(initial) = 0
    v(final) = ?

    We are given E in V/m, q (the charge of the particle) in C, T in degrees, L in m, and mass in kg.


    2. Relevant equations
    What I used:
    Conservation of energy: [tex]K_i+U_i=K_f+U_f[/tex] (1)
    Non-relativistic kinetic energy of a mass: [tex]K=\frac{mv^2}{2}[/tex] (2)
    Change in potential in a uniform electric field: [tex]\Delta\phi=-Ed[/tex] (3)
    Electric potential energy: [tex]U=q\phi(x)[/tex] (4)
    Potential difference: [tex]\Delta\phi=\phi_i - \phi_f[/tex] (5)

    Definition of vars:
    U = electric potential energy
    [tex]\phi[/tex] = electric potential
    K = kinetic energy
    m = mass
    v = velocity
    q = charge
    d = displacement
    E = electric field

    3. The attempt at a solution
    First, I solved (1) for final kinetic energy and substituted (2) and (4) into (1):
    [tex]q\phi_i - q\phi_f = \frac{mv^2}{2}[/tex]

    Factoring out the q and noting (5):
    [tex]q\Delta\phi = \frac{mv^2}{2}[/tex]

    Solving for the wanted variable v:
    [tex]v = \sqrt{\frac{2q\abs{\Delta\phi}}{m}}[/tex]

    Noting (3):
    [tex]v = \sqrt{\frac{2qEd}{m}}[/tex]
    Call this equation (6).

    We see that we need to find the displacement. Electric force is conservative, so I take the direct vector from initial position to final. Looking at the x and y components separately, we see that the y component is perpendicular to the electric field, so no work is done to move along the y component. We need only look at the x component. To find this, I took the pivot point as (0,0).

    The initial point was calculated to be: [tex](L\cos{\theta^'},L\sin{\theta^'})[/tex], where theta prime is the angle between the string at initial position to the y axis, or 90 minus the given theta.

    The final point (parallel to E) was found to be (L, 0).

    Taking into account only x as described previously, we see that the total displacement we care about is: [tex]L\cos{\theta^'}-L[/tex]

    Substituting the final displacement formula into (6):
    [tex]v = \sqrt{\frac{2qE(L\cos{\theta^'}-L)}{m}}[/tex]

    This yields v = .135 m/s, whereas the answer is v = .167 m/s. Can you tell me where I went wrong?

    Note: I've also used the following displacement formulas. These were also wrong, but by a far larger margin. (A little more than 2x the right answer)

    [tex]d = 2L^2+2L^2\cos{\theta}[/tex]
    (Derived from Euclid's general form for the Pythagorean theorem)

    [tex]d = \frac{T2\pi L}{360}[/tex]




    Random footnote: The forum appends the template onto the end of my post every time I hit Preview Post. Bug because I'm using Chrome, or something else?
     
  2. jcsd
  3. Sep 25, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I don't understand the business with θ' instead of θ. If the string is at an angle θ with the +x axis (where the field points), then the displacement along x is just L(1 - cosθ).
     
  4. Sep 25, 2009 #3
    You're absolutely right. I don't know why, but I got it in my head to use the other angle. For kicks, after you gave me the right answer, I replaced the cos(theta') with sin(theta') and got the right answer as well. Thank you for the help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Particle around pivot in uniform E
Loading...