Calculating Work for a Bead on a Square Loop in an Electric Field

[scorpius1782]

Homework Statement

A small bead is on square loop in the xz plane with dimensions (±R, 0, ±R). An electric field is turned that is ##\vec{E}(\vec{r})=-Cx\hat{z}##

Calculate W

Homework Equations

##W= \int q\vec{E}\bullet dl##

The Attempt at a Solution

Starting with z direction:

[tex]q\int_{-R}^R -Cx\hat{z} (dx\hat{x} + dy\hat{y}+ dz\hat{z})[/tex]

=##-Cq \frac {x^2}{2} |_{-R}^R=-CqR^2##

I believe that the total work would just be 2 times this quantity because since the E field isn't in the correct direction in order to move the charge in the x direction. But this doesn't make sense to me within the problem context. Am I missing something?

 

[tiny-tim]

hi scorpius1782!

(type \cdot not \bullet )

[tex]q\int_{-R}^R -Cx\hat{z} (dx\hat{x} + dy\hat{y}+ dz\hat{z})[/tex]

sorry, but that makes no sense

you need to write a dot product

try again (and use a specific arm of the loop) ​

 

[scorpius1782]

thank you for the \cdot! I couldn't remember what it was (kept thinking \vdot) so I gave up and used bullet.
the vertical (z) part of the box from -R to R at x=-R
##\int_{-R}^R-Cqx\hat{z}\cdot dz \hat{z}##

I didn't think of it before but since here x=-R I can put that in for x in this integral. I believe that since there is no change in x or y then dx and dy are zero.

This means that ##CqR\int_{-R}^R\hat{z}\hat{z}dz=CqR(R+R)=2CqR^2##

Then for x=R: ##\int_{R}^{-R}-CqR\hat{z}\cdot dz \hat{z}##
##=-CqR\int_{R}^{-R}\hat{z}\hat{z}dz=-CqR(-R-R)=-2CqR^2##

Is this better?

 

[tiny-tim]

… x=-R …

This means that ##CqR\int_{-R}^R\hat{z}\hat{z}dz=CqR(R+R)=2CqR^2##

yes

Then for x=R: ##\int_{R}^{-R}-CqR\hat{z}\cdot dz \hat{z}##
##=-CqR\int_{R}^{-R}\hat{z}\hat{z}dz=-CqR(-R-R)=-2CqR^2##

erm …

 

[scorpius1782]

Hmm, seems I didn't distribute my negative sign there!

 

[tiny-tim]

are you in the loop now?

 

[scorpius1782]

I am, thank you. I assume since you made no comment that my understanding of the horizontal forcei is correct? That is where z=±R there is no work?

 

[tiny-tim]

yes, for displacement of constant z, the component of the field along the wire is zero, so the work done is zero

 

[scorpius1782]

Thanks for the help!