- #1

- 7

- 0

## Main Question or Discussion Point

Hey guys and gals, this isn't actually an assignment of any sort, so I didn't want to put it in the homework section. This is also my first post, though I have been lurking for quite a while, reading the copious amounts of information available here. :p

Anyhow, could somebody please elaborate on why it doesn't seem to work to treat the variables as a constant even though they are seemingly independent? As demonstrated below, I've tried it on two problems and both times it gives the wrong answer, though I am not sure why. I think I'm probably missing some intuition for these line integrals, and would very much appreciate a helping hand. Thanks!

For the field [tex]\vec{E}= 54y \mathbf{i} + 54x \mathbf{j} + 3 \mathbf{k}[/tex] find the potential function [tex]V_E(x,y,z)[/tex].

[tex]V_E(x,y,z) = -\int \vec{E}\bullet \vec{ds}

= - \int (54y \mathbf{i} + 54x \mathbf{j} + 3 \mathbf{k}) \bullet (dx\mathbf{i} + dy \mathbf{j}+ dz\mathbf{k})

= - \int (54y\ dx + 54x\ dy + 3\ dz)[/tex]

[tex]\therefore V_E(x,y,z)= - 108xy - 3z + C[/tex]

But according to the book, the answer is actually [tex]V_E(x,y,z) = -54xy - 3z[/tex]

***

Suppose our field [tex]\vec{E} = \dfrac{F}{a^2}(yz \mathbf{i} + xz \mathbf{j} + xy \mathbf{k})[/tex], where F and a are constants. Calculate the potential difference [tex]\Delta V_E[/tex] between the origin and the point (1,1,1) along the straight line given by the formula (s, s, s), where the variable s runs from 0 to 1.

Using the same method as above, [tex]V_E(x,y,z) = - \dfrac{F}{a^2}\int (yz\ dx + xz\ dy + xy\ dz) = - \dfrac{3F}{a^2}(xyz) + C[/tex]

[tex]\Rightarrow \Delta V_E = V_E(1,1,1) - V_E(0,0,0) = - \dfrac{3F}{a^2}[/tex]

However, this is wrong, according to the book, and is out by a factor of three. Using a different method, spotting that along the line, x=y=z, leads to a different integral for the potential: [tex]V_E(x,y,z) = - \dfrac{F}{a^2}\int (xx\ dx + xx\ dx + xx\ dx) = - \dfrac{F}{a^2}\int (3x^2\ dx) = - \dfrac{F}{a^2}x^3[/tex]

[tex]\therefore \Delta V_E = - \dfrac{F}{a^2}[/tex]

This is the correct answer, according to the book.

Anyhow, could somebody please elaborate on why it doesn't seem to work to treat the variables as a constant even though they are seemingly independent? As demonstrated below, I've tried it on two problems and both times it gives the wrong answer, though I am not sure why. I think I'm probably missing some intuition for these line integrals, and would very much appreciate a helping hand. Thanks!

**Problem 1**For the field [tex]\vec{E}= 54y \mathbf{i} + 54x \mathbf{j} + 3 \mathbf{k}[/tex] find the potential function [tex]V_E(x,y,z)[/tex].

[tex]V_E(x,y,z) = -\int \vec{E}\bullet \vec{ds}

= - \int (54y \mathbf{i} + 54x \mathbf{j} + 3 \mathbf{k}) \bullet (dx\mathbf{i} + dy \mathbf{j}+ dz\mathbf{k})

= - \int (54y\ dx + 54x\ dy + 3\ dz)[/tex]

[tex]\therefore V_E(x,y,z)= - 108xy - 3z + C[/tex]

But according to the book, the answer is actually [tex]V_E(x,y,z) = -54xy - 3z[/tex]

***

**Problem 2**Suppose our field [tex]\vec{E} = \dfrac{F}{a^2}(yz \mathbf{i} + xz \mathbf{j} + xy \mathbf{k})[/tex], where F and a are constants. Calculate the potential difference [tex]\Delta V_E[/tex] between the origin and the point (1,1,1) along the straight line given by the formula (s, s, s), where the variable s runs from 0 to 1.

Using the same method as above, [tex]V_E(x,y,z) = - \dfrac{F}{a^2}\int (yz\ dx + xz\ dy + xy\ dz) = - \dfrac{3F}{a^2}(xyz) + C[/tex]

[tex]\Rightarrow \Delta V_E = V_E(1,1,1) - V_E(0,0,0) = - \dfrac{3F}{a^2}[/tex]

However, this is wrong, according to the book, and is out by a factor of three. Using a different method, spotting that along the line, x=y=z, leads to a different integral for the potential: [tex]V_E(x,y,z) = - \dfrac{F}{a^2}\int (xx\ dx + xx\ dx + xx\ dx) = - \dfrac{F}{a^2}\int (3x^2\ dx) = - \dfrac{F}{a^2}x^3[/tex]

[tex]\therefore \Delta V_E = - \dfrac{F}{a^2}[/tex]

This is the correct answer, according to the book.