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

AI Thread Summary
The discussion focuses on calculating the electric potential difference between two points in a given electric field, E = 2xi - 3y² j N/C. The user initially calculated the potential difference incorrectly, obtaining -10 V instead of the correct +6 V, due to a sign error in the integration process. After guidance, the user corrected the mistake by properly evaluating the integrals and recognizing the importance of using parentheses to avoid confusion with negative signs. The conversation also clarifies that the z-coordinate for point rA is zero, which is irrelevant since the electric field has no z-component. Ultimately, the user successfully arrives at the correct potential difference after addressing the calculation errors.
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 = -\int_A^B E\cdotds

The Attempt at a Solution



ΔV= -\int_1^22x dx - \int-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:

- \left[ x^2 \right]_{1}^{2} - \left[ -y^3 \right]_{-2}^{1} =
-(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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