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Work and Electric Potential

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

    Two parallel metal plates 5.0 cm apart have a potential difference between them of 75 V. The electric force on a positive charge of 3.2*10^-19 C at a point midway between the plates is approximately

    a. 1.6*10^-18 N
    b. 2.4*10^-17 -
    c. 4.8*10^-18
    d. 4.8*10^-16
    e. 9.6*10^-17

    2. Relevant equations

    V = E*d = (F/q)*d

    3. The attempt at a solution
    F = (V*q)/d = (75 V)*(3.2*10-19 C)/(0.025 m) = 9.6*10^-16 N???
    But this does not match any of the choices . . .except for being close to E.

    Is my arithmetic wrong or is the question in need of editing?

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

    The electric potential in a region of space is given by V(x,y,z)=10 (V/m)x +20 V/m(y) + 30(V/m)z. The z-component of the electric field in this region is

    a. (-30 V/m)*k vector
    b. (-10 V/m)*i vector
    c. (20 V/m)*j vector
    d. (-20 V/m)*j vector
    e. (30 V/m)*k vector

    2. Relevant equations

    See below.

    3. The attempt at a solution

    Is the answer e. (30 V/m)*k vector since V/m = N/C (both units for electric field)?

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

    When 5.0 C of charge moves at constant speed along a path between two points differing in potential by 12 V, the amount of work done by the electric field is

    a.2.4 J
    b. 0.42 J
    c. 5.0 J
    d. 12 J
    e. 60 J

    2. Relevant equations

    U = q*V ?? (correct formula?)

    3. The attempt at a solution

    U = 5.0 C*12V = 60 J?

    Last edited: Feb 6, 2007
  2. jcsd
  3. Feb 6, 2007 #2
    For the first one, I got 4.8 * 10^-16....the electric field is constant so the force will be constant no matter what point you choose in the field...looks like you chose the midway point, well the potential is not constant throughout the field so if you choose the midway point, you have to divide the potential by half as well.

    I dont know about 2 and 3....
  4. Feb 6, 2007 #3
    Again, that cannot be electric potential because potential is not a vector.

    Concerning the 1st question, E=V/d since electric field between parallel plates is uniform. Then, force =qE. Thus answer is d.

    The third question however is trickier! (although your answer is correct)

    You see if the field alone were acting on it then the motion would be accelerated. But since charge is moving at uniform velocity then we have to assume that work has to be done by an external agency to keep the charge in the same speed. Thus the work has to be negative. Atleast that is what I conclude!

    So W= -60 J because external force is directed opposite to field.

    So could you please chack again whether the constant speed part is there!
  5. Feb 6, 2007 #4
    For the last one, the constant speed part is included.

    For the second part, based on the partial derivative definition E = -dV/dx, the electric field of the z-component will be -30 V/m*vector k?
  6. Feb 6, 2007 #5
    Does the x, y and z in the equation refer to the axes in free space?!
  7. Feb 6, 2007 #6
    I don't understand?
  8. Feb 7, 2007 #7
    Are x,y and z unit vectors or just scalars in the 3D equation?!
  9. Feb 7, 2007 #8
    V is just a function of x, y, and z in the problem, I think.

    E_z = -dV/dz?
    Last edited: Feb 7, 2007
  10. Feb 7, 2007 #9
    Then the answer would be given by differentiating the equation with respect to z and put a negative sign, i.e: E = -dV/dz .

    Thus, answer is -30 V/m
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