I'm trying to find the potential difference of a punctual charge $Q=-3\mu C$ that moves from cartesian coordinate A(1,1,-1) to B(2,2,-1) in the external electric field $$\vec{E}=\left[ \frac{z}{x^2y}, \frac{z}{xy^2}, -\frac{1}{xy}\right].$$

So far I tried this:

\begin{align*} V_{ba} &= - \int_a^b \vec{E} \cdot d\vec{L}\\ &=-\int_1^2 \left[ \frac{z}{x^2y}, \frac{z}{xy^2}, -\frac{1}{xy}\right] \cdot \left[dx,0,0 \right]-\int_1^2 \left[ \frac{z}{x^2y}, \frac{z}{xy^2}, -\frac{1}{xy}\right] \cdot \left[0,dy,0 \right]\\ &=\frac{-z}{2y} - \frac{z}{2x}. \end{align*}

I dont know how to interpret this result. Am I making a mistake ?


In each integral you need to consider the path you are taking and what values x, y, and z are taking along those paths.

Assuming you are following the paths in the order of your integrals, first you are moving from x=1 to x=2 when y=1 and z=-1. Therefore, you must plug in those values for y and z in the first integral.

Then the second integral corresponds to moving from y=1 to y=2 when x=2 and z=-1. Therefore you must use those values for x and z in the second integral.

Following this you will then get an actual number for your final answer. Notice that if you had moved in y first and then moved in x that we get the same answer. This is good, since we know the electrostatic force is conservative.

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  • $\begingroup$ So , like this ? $-\int_1^2 \frac{-1}{x^2}dx -\int_1^2 \frac{-1}{y^2}dy $ $\endgroup$ – Liam F-A Mar 2 '18 at 4:28
  • $\begingroup$ Close! Remember in your second integral x=2, since you already moved from 1 to 2 in the x-direction. $\endgroup$ – BioPhysicist Mar 2 '18 at 12:30

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