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

AI Thread Summary
The discussion revolves around calculating the electric force on an electron in a varying electric potential within an xy plane. The user attempted to derive the electric field components using partial derivatives of the potential, resulting in values of 1250 N/C for the x-component and -250 N/C for the y-component. They then calculated the magnitude of the electric field and used the formula F=qE to find the force on the electron, but indicated that their answer was incorrect. A clarification about the source of the potential values led to a realization of a mistake in interpretation. The thread emphasizes the importance of accurately assessing the electric potential and its derivatives to determine the correct electric force.
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.
 
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