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Electron accelerated through potential

  1. Nov 21, 2005 #1
    I have a very basic question. An electron is accelerated through a potential [tex]V[/tex], what is the velocity? Obviously, this can be solved using conservation of mechanical energy, but why am I off by a negative?

    [tex]K_i + U_i = K_f + U_f [/tex]
    [tex] 0 + (-qV) = \frac{1}{2} m v^2 + 0 [/tex]
    [tex] v = \sqrt{-2qV/m} [/tex]

    My problem must be in the initial potential energy. But I do not see how, because the potential difference is positive and the charge (electron) is negative.
     
    Last edited: Nov 21, 2005
  2. jcsd
  3. Nov 21, 2005 #2

    mezarashi

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    The energy (Vq) should be positive, as the change potential is positive from + to - which is the case for a hole (+q), but an electron (-q) moves from - to +.
     
  4. Nov 21, 2005 #3
    I see. So, one must consider the change in potential relative to where it starts and where it ends.

    Does this mean that electric potential energy must always be written [tex]\Delta U[/tex] never [tex]U_f, U_i[/tex]?
     
  5. Nov 22, 2005 #4

    mezarashi

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    Well, whether you write it in delta notation or Uf - Ui makes no difference. The difference is the difference.

    The point to notice is being consistent. For example, if you draw the electric field lines to go from positive charge to negative charge. The positive direction indicates the movement of a positive charge in the field, then the opposite holds for negative charges. Voltage can be defined as the negative of the integral of this E-field over some distance:

    [tex] V = -\int\limits_{A}^{B} E \cdot dx [/tex]

    or the non-calculus version just incase:
    [tex] V = - E \cdot \Delta x [/tex]

    [tex] \Delta V = V_f - V_i = \frac{\Delta U}{q} = \frac{U_f - U_i}{q}[/tex]

    So how consistent you are with which direction the charge is moving along the field will be important in making sure you get your signs right.
     
    Last edited: Nov 22, 2005
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