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Electric Field direction?

  1. Nov 11, 2007 #1
    1. The problem statement, all variables and given/known data

    7. In Fig. 28-35, an electron accelerated from rest through potential difference [tex]{V}_{1} = 1.00{\textcolor[rgb]{1.00,1.00,1.00}{.}}kV[/tex] enters the gap between two parallel plates having separation [tex]d = 20.0{\textcolor[rgb]{1.00,1.00,1.00}{.}}mm[/tex] and potential difference [tex]{V}_{2} = 100{\textcolor[rgb]{1.00,1.00,1.00}{.}}V[/tex]. The lower plate is at the lower potential. Neglect fringing and assume that the electron’s velocity vector is perpendicular to the electric field vector between the plates. In unit-vector notation, what uniform magnetic field allows the electron to travel in a straight line in the gap.

    [​IMG]

    2. Relevant equations

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

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

    [tex]
    {\vec{F}_{B}} = q{\vec{v}}{\times}{\vec{B}}
    [/tex]

    3. The attempt at a solution

    Ok, this problem does not seem that hard, however the wording isn’t really clear. So let me see if I understand this problem, you have an electron being accelerated by [tex]{V}_{1}[/tex] in to an electric field produced by two plates. Where the bottom plate has potential [tex]{V}_{2}[/tex].

    Ok, so my question is if the bottom plate has potential [tex]{V}_{2}[/tex], does the top plate have a potential?

    Also, what about the electric field between the two plates how can you tell in what direction the electric field points in?

    Any help is appreciated.

    Thanks,

    -PFStudent
     
    Last edited: Nov 11, 2007
  2. jcsd
  3. Nov 12, 2007 #2
    Hey,

    So, I was looking at this problem again and I am still not all that sure about how you can tell which way the electric field points? How is that you determine that in this problem?

    Any help is appreciated.

    Thanks,

    -PFStudent
     
  4. Nov 12, 2007 #3

    Doc Al

    User Avatar

    Staff: Mentor

    The electric field points from higher to lower potential, which can be seen from this equation of yours (that's the meaning of the minus sign):
     
  5. Nov 26, 2007 #4
    Hey,

    I was looking at this problem again and had a quick question. When finding the potential difference,

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

    Do you always integrate from the lower potential to the higher potential? In other words when finding the Electric field would I integrate from,

    [tex]
    {r_{0}} = {-{\frac{d}{2}}}
    [/tex]

    to

    [tex]
    {r_{1}} = {+{\frac{d}{2}}}
    [/tex]

    So, that the integral would look like the following,

    [tex]
    {\Delta{V}_{2}} = {-}{\int_{{-{\frac{d}{2}}}}^{{{\frac{d}{2}}}}}{|{\vec{E}}_{}(r)|}{}{|d{{\vec{r}}}|}{cos\theta}
    [/tex]

    Where, [tex]{\theta} = {180}^{o}[/tex], since [itex]\vec{E}[/itex] points downward (from the higher potential to the lower potential) and [itex]d{\vec{r}}[/itex] points upward from [itex]{r_{0}}[/itex] to [itex]{r_{1}}[/itex].

    Is that right?

    Thanks,

    -PFStudent
     
    Last edited: Nov 26, 2007
  6. Nov 28, 2007 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Note that you are always finding the potential of r_1 with respect to r_0. If you reverse the integration (integrating from r_1 to r_0) the sign changes:

    [tex]\int_a^b Fdx = -\int_b^a Fdx[/tex]

    So it doesn't really matter which way you integrate. What does matter is the direction of [itex]\vec{E}[/itex] compared to [itex]d\vec{r}[/itex].
     
  7. Nov 28, 2007 #6
    Hey,

    Forgot to say thanks for the above info. That information had helped me better understand how to solve these type of problems. Thanks.

    Ah...yes I do remember that. Thanks for reminding me.

    Ok, so then on the below integral,

    Do I have the limits correct?

    Ok, so was I correct in performing the integration the way I did, Where [itex]{\theta} = {180}^{o}[/itex], since [itex]\vec{E}[/itex] points downward (from the higher potential to the lower potential) and [itex]d{\vec{r}}[/itex] points upward from [itex]{r_{0}}[/itex] to [itex]{r_{1}}[/itex]?

    Also, since you mentioned that propery of integrals--I could alternatively intregrate the following way,

    [tex]
    {\Delta{V}_{2}} = {}{\int_{{{\frac{d}{2}}}}^{{{-}{\frac{d}{2}}}}}{|{\vec{E}}_{}(r)|}{}{|d{{\vec{r}}}|}{cos\theta}
    [/tex]

    Is the above integral correct also?

    So, in the end was the below integral correct?

    [tex]
    {\Delta{V}_{2}} = {-}{\int_{{-{\frac{d}{2}}}}^{{{\frac{d}{2}}}}}{|{\vec{E}}_{}(r)|}{}{|d{{\vec{r}}}|}{cos\theta}
    [/tex]

    Thanks so much for the help,

    -PFStudent
     
  8. Nov 28, 2007 #7

    Doc Al

    User Avatar

    Staff: Mentor

    Yes. You are finding the potential difference of r_1 with respect to r_0.



    Absolutely.

    Yes indeed.


    You bet.
     
  9. Nov 29, 2007 #8
    Hey,

    Thanks for the help Doc Al! :smile:

    Thanks,

    -PFStudent
     
    Last edited: Nov 29, 2007
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