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

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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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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
 
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:
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:
 
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.
 
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