How Is Electric Potential Calculated Between Two Points in a Field?

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Homework Help Overview

The discussion revolves around calculating the change in electric potential between two points in an electric field defined by E = 2xi - 3y²j N/C. The original poster attempts to find the potential difference from position rA = i - 2j m to rB = 2i + i + 3k m, using the integral of the electric field.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the integral for potential difference, noting discrepancies in results and questioning the correctness of the integral setup and limits. There is also mention of potential sign errors in the calculations.

Discussion Status

Some participants have provided guidance on identifying possible errors in the calculations, particularly regarding the handling of signs in the integral. There is an ongoing exploration of the implications of the z-coordinate in relation to the electric field's components.

Contextual Notes

Participants note the absence of a z-coordinate for rA and discuss its relevance given that the electric field has no z-component. This raises questions about the dimensionality of the problem and its impact on the calculations.

Strawer
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Homework Statement



An electric field is given by E = 2xi-3y2j N/C. Find the change in potential from the position rA = i - 2j m to rB = 2i + i + 3k m.

Homework Equations



VB - VA = -[itex]\int_A^B[/itex] E[itex]\cdot[/itex]ds

The Attempt at a Solution



ΔV= -[itex]\int_1^2[/itex]2x dx - [itex]\int[/itex]-3y^2 dy

The second integral is supposed to be from -2 to 1

And when I calculate this I get -10 V when the answer is +6 V

I also noticed that if I simply invert the limits on the integral I get +6 V, is it just a coincidence or have I calculated the integral wrong?
 
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You can have a sign error somewhere. Show your work. ehild
 
Well i have noticed that I'm very prone to making rudimentary mistakes often but I don't believe that to be the case here.

When calculating the integral I do as follows:

- [itex]\left[ x^2 \right]_{1}^{2}[/itex] - [itex]\left[ -y^3 \right]_{-2}^{1}[/itex] =
-(4-1) - (-1+8)= -3 - 7 = -10 V
 
You made the sign error at y^3. At the lower bound it is -(-(-(-8)))

-x^2-(-y^3)=-x^2+y^3.


-(4-1)+(1^3-(-2)^3)=-3+(1+8)=6

ehild
 
ehild said:
You made the sign error at y^3. At the lower bound it is -(-(-(-8)))

-x^2-(-y^3)=-x^2+y^3.


-(4-1)+(1^3-(-2)^3)=-3+(1+8)=6

ehild

Oh I see now. My homework would be a lot loss painful if I could avoid stuff like this but it doesn't matter how thorough I am, I often miss things anyways. I think I might have a mild form of dyscalculia or something..

But thank you for your assistance!
 
Last edited:
Get rid of the lot of minuses as soon as possible and use parentheses. No dyscalculia. I got the correct result at the fourth attempt.:smile:
 
ehild said:
Get rid of the lot of minuses as soon as possible and use parentheses. No dyscalculia. I got the correct result at the fourth attempt.:smile:

Hehe well it's good to know that I'm not the only one who struggles with stuff like that :smile:
 
What is the z coordinate for rA ?
 
SammyS said:
What is the z coordinate for rA ?

There isn't a z coordinate for rA but as I understand that doesn't matter anyways since the electric field has no component in the z direction.
 
  • #10
Strawer said:
There isn't a z coordinate for rA but as I understand that doesn't matter anyways since the electric field has no component in the z direction.

That is the z coordinate is 0. :smile:

ehild
 
  • #11
ehild said:
That is the z coordinate is 0. :smile:

ehild
Yup , so rB - rA has a z component.
 
  • #12
It has, but E has not.

ehild
 

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