Potential Difference from Given Efield

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Homework Help Overview

The problem involves calculating the potential difference from the origin to a point (2, 2) m in a given electric field described by E=2x^2 i +3y j, with units in V/m. The relationship between electric field and potential is noted, but there is confusion regarding the integration process and the resulting values.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of the electric field components to find the potential difference, with one participant expressing uncertainty about the correctness of their calculations and the discrepancy between their result and the answer guide.

Discussion Status

There is an ongoing exploration of the integration process and the potential for errors in the answer guide. Participants are questioning the accuracy of the electric field provided and whether the integration was performed correctly.

Contextual Notes

Participants are considering the possibility of errors in the provided electric field and the answer guide, which may affect the interpretation of the problem.

SMA777
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Homework Statement


The electric field in a region is given by E=2x^2 i +3y j where the units are in V/m. What is the potential difference from the origin to (x, y) = (2, 2) m?

Homework Equations



E = -gradient of V


The Attempt at a Solution



- derivative of the x-component: 2/3 x^2 from 0 to 2 = -16/3 x
- derivative of y-component: 3/2y^2 from 0 to 2 = -6 y

And then I added them for −11&1/3 V
But this wrong. I'm shown the correct answer is -27 & 1/3 V

Any tips on how that could be? Thanks !
 
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SMA777 said:
- derivative of the x-component: 2/3 x^2 from 0 to 2 = -16/3 x
- derivative of y-component: 3/2y^2 from 0 to 2 = -6 y

And then I added them for −11&1/3 V
But this wrong. I'm shown the correct answer is -27 & 1/3 V

Any tips on how that could be? Thanks !

It was integration, and the integral of x^2 is x^3/3, and there is no x or y after substituting x=2 and y=2.

The result is correct however.

ehild
 
Oh ,that's right! I meant integrated, sorry.

Wait, but my answer guide says -27 & 1/3 V and the sum of my answers is −11&1/3 V ? So, how do I got from my integration to the -27 & 1/3 V? I tried just adding and, as you saw, got −11&1/3 V which was incorrect
 
The answer guides are wrong sometimes. Check if you copied the electric field correctly.

ehild
 

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