(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

As a rough measure of the relativistic "flattening" of the configuration of electric field lines from a moving charge, we might use the angle [itex]\alpha[/itex] between two conical surfaces which include between them half the total electric flux. That is, half the flux through a sphere shall be contained in the equatorial zone between [itex]\theta[/itex]' = [itex]\frac{\pi}{2}[/itex]+[itex]\frac{\alpha}{2}[/itex] and [itex]\theta[/itex]' = [itex]\frac{\pi}{2}[/itex]-[itex]\frac{\alpha}{2}[/itex]. Consider only the extreme relativistic case, with [itex]\gamma[/itex]>>1. Then only angles [itex]\theta[/itex]' such that [itex]\theta[/itex]'=[itex]\frac{\pi}{2}[/itex]-[itex]\epsilon[/itex], with [itex]\epsilon[/itex]<<1 need be considered. Show first that Eq.12 can be approximated, in this case, as

E' = [itex]\frac{Q}{(r')^{2}}[/itex][itex]\frac{\gamma}{(1+\gamma^{2}\epsilon^{2})^{3/2}}[/itex]

Now let [itex]\epsilon[/itex] range from [itex]\frac{-\alpha}{2}[/itex] to [itex]\frac{+\alpha}{2}[/itex] and integrate to obtain the flux through the narrow equatorial belt.

2. Relevant equations

Eq. 12: E' = [itex]\frac{Q}{(r')^{2}}[/itex][itex]\frac{1-\beta^{2}}{(1-\beta^{2}sin^{2}\theta')^{3/2}}[/itex]

[itex]\gamma[/itex]=[itex]\frac{1}{\sqrt{1-\beta^{2}}}[/itex]

Gauss' Law: Electric flux out of a sphere = 4[itex]\pi[/itex]Q

(My book uses Gaussian units)

3. The attempt at a solution

I'm still stuck on approximating equation 12. I have tried the small-angle approximation for sin[itex]^{2}[/itex]([itex]\frac{\pi}{2}[/itex]-[itex]\epsilon[/itex]), and I have tried converting this to cos[itex]^{2}[/itex]([itex]\epsilon[/itex]) and approximating with a second order Taylor series, and I've tried expanding the entire equation, but I can't make Equation 12 agree with the equation they gave me.

This is from Purcell's book, usually used for honors freshman physics, so the math probably isn't terribly advanced, but I can't see what else to try.

I have also tried skipping that part and attacking the integral directly, but I can't seem to make it into something I can do by hand or look up in a table.

For the integral, I know that E' is radial, so [itex]\vec{E'}[/itex][itex]\cdot[/itex][itex]\vec{dA}[/itex] will simply equal E'dA. So to find dA in terms of r' and [itex]\epsilon[/itex] I figure that the area of a thin ring of the sphere near the equator is the width r'd[itex]\epsilon[/itex] times the circumference of the ring 2[itex]\pi[/itex](r'cos[itex]\epsilon[/itex]).

So I get the integral:

[itex]\int[/itex][itex]^{\alpha/2}_{-\alpha/2}[/itex][itex]\frac{Q\gamma*2\pi*(r')^{2}cos\epsilon}{(r')^{2}(1+\gamma^{2}\epsilon^{2})^{3/2}}[/itex]d[itex]\epsilon[/itex]

And I set that equal to half the total flux through the sphere, 2[itex]\pi[/itex]Q and simplify a bit to get:

[itex]\frac{1}{\gamma}[/itex] = [itex]\int[/itex][itex]^{\alpha/2}_{-\alpha/2}[/itex][itex]\frac{cos\epsilon}{(1+\gamma^{2}\epsilon^{2})}[/itex]d[itex]\epsilon[/itex]

But I don't know how to evaluate that integral. So someone please let me know if I have the setup wrong or if I just need an integral trick I don't know.

The book provides the final answer by the way, and it is:

[itex]\alpha[/itex] = [itex]\frac{2}{\sqrt{3}\gamma}[/itex]

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# Homework Help: Measure of Relativistic Flattening of Electric Field

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