What Is the Electric Force on an Electron in a Varying Electric Potential?

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Homework Help Overview

The problem involves determining the electric force on an electron placed in an xy plane where the electric potential varies with x and y. The potential is defined with a vertical scale of 500 V, and the task is to express the electric force in unit-vector notation.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss evaluating the electric field components using partial derivatives of the potential. There is an attempt to calculate the electric field and subsequently the force on the electron. Some participants question the values used for the potential differences and the interpretation of the graph.

Discussion Status

The discussion is ongoing, with participants providing insights into potential errors in calculations and clarifying the interpretation of the provided figures. There is recognition of mistakes in the initial approach, and some participants express understanding of their errors.

Contextual Notes

There are references to specific values derived from the graph, and some participants express uncertainty about the accuracy of these values. The discussion includes attempts to share visual aids to clarify the problem context.

w3dnesday
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Homework Statement



An electron is placed in an xy plane where the electric potential depends on x and y as shown in the figure (the potential does not depend on z). The scale of the vertical axes is set byVs = 500 V. In unit-vector notation, what is the electric force on the electron?
https://www.facebook.com/photo.php?fbid=1403786173283706&set=a.1403786203283703.1073741827.100009570105613&type=1

Homework Equations

The Attempt at a Solution


I tried evaluating the graph and seeing as it is a straight line for both the x and y components of the E field i was thinking i could get a partial derivative
∂V/∂x=-(-500v/0.4m)=1250N/C
∂V/∂y=-(100V/.4m)=-250N/C
I think this is where i went wrong.
anyhow, I went on to find the magnitude and direction of the E field
|E|=√(1250^2+250^2)=1274.75N/C
θ=tan^-1(-250/1250)=-11.31°
the next step given the force on an electron as F=qE
finding this for the separate components in vector notation give q=1.6*10^-19C
so F=(1.6*10^-19C)*[(1250N/Ci), (-250N/Cj)]
it is not the right answer so my understanding is failing at a point
thankyou
 
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Including the picture like this does not seem to work, but this link should work.
w3dnesday said:
∂V/∂x=-(-500v/0.4m)=1250N/C
∂V/∂y=-(100V/.4m)=-250N/C
How did you get those 500 V and 100 V? The difference between 0 and 0.4 m looks larger in both pictures.
 
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oh... derpy me thank you for pointing out my mistake... I know what i did now... i thought that was were i went wrong but i just could not see what it was
 
Or maybe this way:

https://scontent-ord1-1.xx.fbcdn.net/hphotos-xta1/v/t1.0-9/11391362_1403786173283706_627839923534487902_n.jpg?oh=61df4a57a6442947a541c4582b4e1ad1&oe=562D9CDD
... so all can see it.

Added in Edit:
It might be helpful to explain what I did to make this image visible:

I used the Link provided by mfb .
I'll speculate that mfb looked into the Original Post someway in order to obtain the link.​

w3dnesday said:

Homework Statement



An electron is placed in an xy plane where the electric potential depends on x and y as shown in the figure (the potential does not depend on z). The scale of the vertical axes is set byVs = 500 V. In unit-vector notation, what is the electric force on the electron?
[ IMG]https://www.facebook.com/photo.php?...703.1073741827.100009570105613&type=1[/PLAIN]
...
I right-clicked on the image which appeared when I followed the link and chose "Copy Image"

Then I simply "Paste"-d the image into the message editor .

An alternative would have been to copy the image to a file on my computer, then use the upload feature at the bottom of the message editor. This method would have the advantage of ensuring that the image will be accessible even if the original image gets deleted or moved or whatever. -- but I was in a hurry and/or was being stingy with my disk space.
 
Last edited:

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