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Electric Field direction?

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Homework Statement



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.

http://img228.imageshack.us/img228/4953/platepicgl6.jpg [Broken]

Homework 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]

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 by a moderator:

Answers and Replies

  • #2
170
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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
 
  • #3
Doc Al
Mentor
44,871
1,119
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):
[tex]
{\Delta{V}_{p}} = {-}{\int_{r_{0}}^{r_{1}}}{\vec{E}_{p1}(r)}{\cdot}{d{\vec{r}}}
[/tex]
 
  • #4
170
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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:
  • #5
Doc Al
Mentor
44,871
1,119
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?
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].
 
  • #6
170
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Hey,

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

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

Ok, so then on the below integral,

[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].
Do I have the limits correct?

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].
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
 
  • #7
Doc Al
Mentor
44,871
1,119
Ok, so then on the below integral,
[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].

Do I have the limits correct?
Yes. You are finding the potential difference of r_1 with respect to r_0.



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]?
Absolutely.

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?
Yes indeed.


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]
You bet.
 
  • #8
170
0
Hey,

Thanks for the help Doc Al! :smile:

Thanks,

-PFStudent
 
Last edited:

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