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Electric field and potential

  1. Mar 2, 2007 #1
    complete data:
    two large horizontal plates are separated by 4mm .the lower plate is at a potential of -6v.what potential should be applied to upper plate to create an electric field strength of 4000 v/m upward in space between the plates
    relavant equation:
    attempt at question:
    but the answer of this question is -22v
    how it will be solved
  2. jcsd
  3. Mar 2, 2007 #2


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    Umm....I'm not sure what your background is, because this is not advanced physics, yet you have posted it in the advanced physics forum. If you know calculus, then just start from the definition:

    [tex] \mathbf{E} = -\nabla V = -\mathbf{\hat{x}} \frac{dV}{dx} [/tex]

    Where the last part applies only in this 1-dimensional case, and I've arbitrarily chosen my coordinate system so that the direction along a perpendicular line from lower plate to upper plate is defined to be the positive x-direction. Since the field is uniform, we can simplify further:

    [tex] |\mathbf{E}| = E = -\frac{\Delta V}{\Delta x} [/tex]

    If you don't know calculus, then ignore what I said before and just start with the formula above. This formula should make sense. The numerator, the change in potential energy per unit charge in moving from one point to another represents the negative of the work done by the electric force over that distance. If you then divide that distance out (the denominator), you just get the electric force per unit charge. Call the voltage on the upper plate V. Then:

    [tex] \Delta V = V - (-6 \ \textrm{V}) [/tex]

    [tex] \Delta x = d = 0.004 \ \textrm{m} [/tex]

    [tex] E = 4000 \ \frac{\textrm{V}}{\textrm{m}} [/tex]

    Solve for V.
    Last edited: Mar 2, 2007
  4. Mar 2, 2007 #3

    Meir Achuz

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    You left out thAt the desired field is UPWARD, so the upper plate shoud be at a lower voltage.
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