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Electric Potential (Why the opposite sign?)

  1. Oct 25, 2007 #1
    1. The problem statement, all variables and given/known data

    Two large, parallel, conducting plates are 12 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electrostatic force of [tex]3.9 \times {10}^{-15} N[/tex] acts on an electron placed anywhere between the two plates. (Neglect fringing.)
    (a) Find the electric field at the position of the electron.
    (b) What is the potential difference between the plates?

    2. Relevant equations

    [tex]
    {\vec{E}}_{p1} = {\frac{\vec{F}_{21}}{q_{2}}}
    [/tex]

    [tex]
    {E_{p1}} = {\frac{|\vec{F}_{21}|}{|q_{2}|}}
    [/tex]

    [tex]
    {E} = {-{\frac{\Delta{V}}{\Delta{r}}}}
    [/tex]

    3. The attempt at a solution

    At first I thought this was a pretty straight-forward problem and got part (a) right but did not understand why part (b) was wrong.

    (a)

    [tex]
    {F_{21}} = 3.9 \times {10}^{-15} N
    [/tex]

    [tex]
    q_{2} = -e
    [/tex]

    [tex]
    {E_{p1}} = {\frac{|\vec{F_{21}}|}{|q_{2}|}}
    [/tex]

    sig. fig. [itex]\equiv[/itex] 2

    [tex]
    {E_{p1}} = 2.4 \times {10}^{4} N/C
    [/tex]

    (b)

    [tex]
    \Delta{r} = 0.12{\textcolor[rgb]{1.00,1.00,1.00}{.}}m
    [/tex]

    [tex]
    {E} = {-{\frac{\Delta{V}}{\Delta{r}}}}
    [/tex]

    [tex]
    {\Delta{V}} = {-}{\Delta{r}}}}{E}
    [/tex]

    sig. fig. [itex]\equiv[/itex] 2

    [tex]
    {\Delta{V}} = -2.9 \times {10}^{3}{\textcolor[rgb]{1.00,1.00,1.00}{.}}V
    [/tex]

    However, the book answer is: [tex]2.9 \times {10}^{3}{\textcolor[rgb]{1.00,1.00,1.00}{.}}V[/tex].

    So, why is the book's answer positive?

    Any help is appreciated.

    Thanks,

    -PFStudent
     
  2. jcsd
  3. Oct 25, 2007 #2
    The key is actually that you found the magnitude of electric field in part one. For the second part you need to use the vector of the electric field, and get the direction right. The only way I know how to tackle it is with calculus, but it is basic so don't freak out.

    [tex]V = - \int_a^b \mathbf{E} \cdot d\matbf{l}[/tex]

    which simplifies (the first one was just for my purposes) to

    [tex] V = - \int_0^{.12} \mathbf{E} \cdot dr\hat{r}[/tex]

    so this is the key step, the one you messed up. If you have your plates drawn on your paper so that the positive one is on top and the negative one is on the bottom the electric field is pointing down. I am integrating in the upwards hat direction, from the negative plate to the positive plate because that is the convention (integrate from negative to positive). So there is an extra minus sign introduced to flip the direction. Hopefully that makes since.

    [tex] V = - \int_0^{.12} -E dr[/tex]

    so then

    [tex] V = E * r[/tex]

    then evaluate with r = .12 and E = your E.
     
  4. Oct 25, 2007 #3

    dynamicsolo

    User Avatar
    Homework Helper

    Just as an aid to remembering the minus sign:

    Keep in mind that the direction of electric fields is defined as the direction that a positive charge carrier will accelerate. Since the potential is the negative integral (or antiderivative) of the electric field*, this means positive charge carriers accelerate from higher to lower potentials. This is consistent with the conservation of mechanical energy, in which kinetic energy increases while (electric) potential energy decreases. Since V = U/q , electric potential energy and electric potential both decrease for accelerating positive charges.

    *This comes from the work-potential energy theorem, which is related to the conservation of mechanical energy discussed a little further on in the paragraph.

    Just the reverse will be the case for electrons and other negative charge carriers. They run "backwards up" the electric field arrows. Since they must also observe conservation of mechanical energy, they will accelerate in the direction of decreasing potential energy, but that now means they accelerate from lower to higher electric potential.

    I harp on this point with my students because people are endlessly victimized on this point on exam questions (especially multiple-choice questions!)...

    P.S. I have to say, though, that the statement of the problem is a bit sloppy: it says nothing about which direction we are to take as positive, so it's not clear what the orientation of the field is to be taken as. I suspect they are just giving the magnitude of the potential difference as the answer...
     
    Last edited: Oct 25, 2007
  5. Oct 27, 2007 #4
    Hey,

    Thanks for the replies Mindscrape and dynamicsolo.

    How I ended up approaching this problem was remembering that when defining the direction of an Electric Field we always use a positive test charge. So, although the charge is negative; I note the electric field lines leave from the postive plate (say on the right) to the negative plate (say on the left) as they normally would. That is, electric field lines leave from the positive plate and terminate on the negative plate. So, although the charge between the two plates is negative the electric field should still point to the left, is that right?

    The displacement vector I reasoned points to the right because that is the direction it would accelerate (towards the postive plate since the charge is negative).

    Now,

    [tex]
    {\Delta{V}_{p}} = {-}{\int_{r_{0}}^{r_{1}}}{\vec{E}_{p1}(r)}{\cdot}{d{\vec{r}}}
    [/tex]

    [tex]
    {\Delta{V}_{p}} = {-}{\int_{r_{0}}^{r_{1}}}{|\vec{E}_{p1}(r)|}{|d{\vec{r}}|}{cos\theta}
    [/tex]

    Since, the vectors are in opposite direction the angle is 180 degrees.

    [tex]
    {\Delta{V}_{p}} = {|\vec{E}_{p1}(x)|}{\int_{x_{0}}^{x_{1}}}{d{{x}}}
    [/tex]

    Also, [itex]{d{x}} > 0[/itex] and [tex]{|\vec{E}_{p1}(x)|} = {{E}_{p1}}[/tex]

    [tex]
    {\Delta{V}_{p}} = {({E}_{p1})}{\Delta{x}}
    [/tex]

    [tex]
    {\Delta{V}_{p}} = {\left(\frac{|\vec{F}_{21}|}{|(-e)|}\right)}{\Delta{x}}
    [/tex]

    [tex]
    {\Delta{V}_{p}} = {\frac{|\vec{F}_{21}|}{e}}{\Delta{x}}
    [/tex]

    [tex]
    {\Delta{V}_{p}} = {\frac{{|\vec{F}_{21}|}{\Delta{x}}}{e}}
    [/tex]

    sig. fig. [itex]\equiv[/itex] 2

    [tex]
    {\Delta{V}_{p}} = 2.9 \times {10}^{3} V
    [/tex]

    Thanks for posting the general form for the potential difference, I forgot I could use that.

    Yea, I think my work above was the approach you described except I integrated from left to right instead of bottom to top (same thing though, right?).

    I was not sure what the convention was for the sign, but your explanation cleared it up a bit. Also, with regards to my approach above--was that the correct approach? Thanks.

    Thanks for the help Mindscrape and dynamicsolo.

    -PFStudent
     
    Last edited: Oct 28, 2007
  6. Oct 27, 2007 #5
    Yep, it looks like you just forgot which direction the electric field went in (positive test charge is the convention), and your new work looks good. You can orient the plates however you feel like, top/bottom, left/right, bottom/top, right/left, angled, etc., as long as you get the vector directions right relative to the positive and negative plate.
     
  7. Mar 5, 2008 #6
    when i try to look at the solution, it says "Physicsforums.com"
     
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