- #1

tylerscott

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## Homework Statement

A charge Q is uniformly distributed with linear density λ over a helix parameterized as [itex]\vec{r}=acos(\theta)\hat{x}+asin(\theta)\hat{y}+ \frac{ h\theta}{2 \pi}\hat{z}[/itex], where a and h are positive constants, and 0<∏<2∏.

a) Find the charge Q

b) Find the electric field on the z-axis at a distance d>h.

## Homework Equations

[itex]\vec{r'}[/itex]=vector to charge element

[itex]\vec{r}[/itex]=vector to test charge

[itex]\vec{E}=\frac{1}{4\pi\epsilon_{O}}\int\frac{(\vec{r}-\vec{r'})}{\left\|\vec{r}-\vec{r'}\right\|^{3}}dq[/itex]

## The Attempt at a Solution

I think I got part a, it's simply the line integral [itex]\int dq=\int\lambda dl=\lambda\int_{0}^{2\pi}\sqrt{a^{2}+\frac{h^{2}}{(2\pi)^{2}}}d \theta =\lambda 2 \pi \sqrt{a^{2}+\frac{h^{2}}{(2\pi)^{2}}} [/itex]

Now part b is the hard one for me. So I defined r' to be the same thing as the parameterized helix:

[itex]\vec{r'}=acos(\theta)\hat{x}+asin(\theta)\hat{y}+ \frac{ h\theta}{2 \pi}\hat{z}[/itex]

and the vector to the point charge as [itex]\vec{r}=d \hat{z}[/itex]

So [itex]\vec{r}-\vec{r'}=-acos(\theta)\hat{x}+-asin(\theta)\hat{y}+ (d-\frac{ h\theta}{2 \pi})\hat{z}[/itex]

And [itex]\left\|\vec{r}-\vec{r'}\right\|=\sqrt{a^{2}+(d-\frac{ h\theta}{2 \pi})^{2}}[/itex]

And the integral will be:

[itex]\vec{E}=\frac{\lambda}{4\pi\epsilon_{O}}\int\frac{(\vec{r}-\vec{r'})}{\left\|\vec{r}-\vec{r'}\right\|^{3}}dl=\frac{\lambda}{4 \pi \epsilon_{O}}\int_{0}^{2\pi}\frac{-acos(\theta)\hat{x}+-asin(\theta)\hat{y}+ (d-\frac{ h\theta}{2 \pi})\hat{z}}{(a^{2}+(d-\frac{ h\theta}{2 \pi})^{2})^{\frac{3}{2}}}d \theta[/itex]

Now, as far as I know, this integral isn't solvable (I've tried separating it into its vector components and solving, but a trig function over a radical isn't exactly a pretty thing to solve. So my first assumption is that I somehow set it up incorrectly.

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