Need help with vector calculue (i think thats wht it is)

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Homework Statement


this is for a physics problem

so formula i need to solve V=1/4\pi\epsilon (Qc/(rc-r \bullet v)
the particle Q moves in a circle of radius a at constant speed w circle is in xy plane at t=0 charge is at (a,0)on x axis. find potentail of charge in z axis.

Homework Equations


The Attempt at a Solution


so i never really know how to find the r the vector position to field point R
I always put like r=(z2+r2) or something but don't know how they find it. For here they did
r= z z^- a(coswt x^ +sinwt y^) than they got
r2=z2 +a2
than they got to something similar to what i always guess r=(z+a)1/2
so how would i figure this out? What is this exactly called so i can maybe find some youtube videos
thanks
 
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nvm i figured it out didnt have to go threw the r=z Z^... i just make a simple graph
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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