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

In summary, the conversation discusses finding the change in potential from two given positions in an electric field. The equation for calculating this change is provided and the attempt at solving it is shown. A sign error is discovered and corrected, leading to the correct result. The z component of the coordinates is also discussed, with the conclusion that it does not affect the calculation as the electric field has no component in the z direction.
  • #1
Strawer
10
0

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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  • #2
You can have a sign error somewhere. Show your work. ehild
 
  • #3
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
 
  • #4
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
 
  • #5
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!
 
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  • #6
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:
 
  • #7
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:
 
  • #8
What is the z coordinate for rA ?
 
  • #9
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
 

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

What is an electric potential integral?

An electric potential integral is a mathematical calculation used to determine the electric potential difference between two points in an electric field. It takes into account the strength and direction of the electric field, as well as the distance between the two points.

How is an electric potential integral calculated?

An electric potential integral is calculated by integrating the electric field strength over the distance between the two points. This can be represented by the equation V = -∫E•ds, where V is the electric potential difference, E is the electric field vector, and ds represents the infinitesimal distance between the two points.

What is the unit of measurement for electric potential integral?

The unit of measurement for electric potential integral is volts (V). This is because the electric potential difference is a measure of the work required to move a unit of charge from one point to another in an electric field.

How is electric potential integral related to electric potential?

Electric potential integral is directly related to electric potential. The electric potential difference, which is calculated using the electric potential integral, is equal to the change in electric potential between two points in an electric field.

What is the significance of electric potential integral in physics?

Electric potential integral is an important concept in physics as it helps us understand the behavior and interactions of charged particles in an electric field. It is used to calculate the electric potential difference, which is essential in determining the direction and magnitude of the electric force experienced by a charged particle in an electric field.

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