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Electron's Speed and Electric Field

  1. Feb 6, 2007 #1
    1. The problem statement, all variables and given/known data

    The voltage between the cathode and the screen of a computer monitor is 12 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed just before it hits the screen?

    a. 8.87*10^7 m/s
    b. 6.5*10^7 m/s
    c. 4.2*10^15 m/s
    d. 7.7* 10^15 m/s
    e. 5.3*10^7 m/s

    2. Relevant equations

    See below.

    3. The attempt at a solution

    q_e = 1.60*10^-19 C
    m_e = 9.11*10^-31 kg

    V = (k*q)/r

    r = (k*Q)/V = (8.988*10^9(1.60*10^-19)/(12,000 v) = 1.1984*10^-13 m

    deltaV = (delta U)/q

    U = deltaV*q = (12,000)*(1.60*10^-19 C)

    F = E/q = (1.92*10^-15 j0/(1.1984*10^-13 m) = 0.01602 N

    a = F/m = (0.0160 N)/(9.11*10^-31 kg) = 1.759*10^28 m/s^2

    v_f^2 = v_i^2 + 2*a*x

    v = sqrt[2*1.7587*10^28 m/s^2*(1.1984*10^-13)] = 6.49*10^7 m/s?????



    1. The problem statement, all variables and given/known data

    A surface is so constructed that, at all points on the surface, the E vector points inward. Therefore, it can be said that


    a.the surface encloses a net positive charge.

    b.the surface encloses a net negative charge.

    c.the surface encloses no net charge.

    d.the surface vector delta S at all points on the surface is necessarily parallel to the electric field vector E.

    e.the surface vector delta S at all points on the surface is necessarily perpendicular to the electric field E.


    2. Relevant equations

    See below.

    3. The attempt at a solution

    By convention, a negative charge corresponds to an inward electric flux while a positive charge corresponds to an outward flux?

    Thanks.
     
    Last edited: Feb 6, 2007
  2. jcsd
  3. Feb 6, 2007 #2
    You're complicating the first answer with too many unknowns.

    You're given P.D and the charge on the particle. What is the work done on the particle at the end of it's trip from the cathode to the screen in terms of these quantities?
     
  4. Feb 6, 2007 #3
    Work done = delta KE

    Work = q*V = 1.192*10^-15 J

    Work = m/2*(v^2)

    sqrt[W/(m/2)] = v_f = sqrt[1.192*10^-15 /(m_e/2)] = 6.49*10^7 m/s

    Well, I'm relieved that my answer checks out.
     
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