Jun 28th 2019, 11:05 PM   #1
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Join Date: Jun 2019
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Hey so I'm kinda really lost in class because my teacher is really confusing. He gave us these three problems for practice for the exam (these aren't assignments or anything).

I'd really appreciate someone breaking down how to do each of these. Thanks so much
Attached Files
 exam 1 example questions.pdf (283.4 KB, 5 views)

Jun 29th 2019, 04:06 AM   #2

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 Originally Posted by pleasehelpmee https://drive.google.com/file/d/1ybz...ew?usp=sharing Hey so I'm kinda really lost in class because my teacher is really confusing. He gave us these three problems for practice for the exam (these aren't assignments or anything). I'd really appreciate someone breaking down how to do each of these. Thanks so much
Surely you have some idea about how to start these. Show us what you have and we'll see where you need the help most. If you don't know how to start, just let us know.

-Dan
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 Jun 29th 2019, 05:01 AM #3 Senior Member   Join Date: Aug 2010 Posts: 434 I presume that you have learned that the force between charge Q and charge Q', with distance r between them is $\displaystyle k_e\frac{QQ'}{r^2}$ where $\displaystyle k_e$ is "Coulomb's constant" (about $\displaystyle 9\times 10^9$ Newton meters squared per Coulomb squared). Imagine dividing the ring into many sectors by drawing lines, a constant angle $\displaystyle \Delta\theta$ apart so that the length of each sector is $\displaystyle R\Delta\theta$ where R is the radius of the ring. Similarly imagine dividing the rod into many short segments of length $\displaystyle \Delta x$. Given point P on the ring, at angle $\displaystyle \theta$ and point Q on the rod at height x, The distance between P and Q is, by the Pythagorean theorem, $\displaystyle r= \sqrt{x^2+ R^2}$. That little segment of the ring is the fraction $\displaystyle \frac{R\Delta\theta}{2\pi R}= \frac{\Delta\theta}{2\pi}$ of the ring so (assuming the charge is uniformly distributed) the charge on it is $\displaystyle \frac{Q\Delta\theta}{2\pi}$. That little segment of the bar is the fraction $\displaystyle \frac{\Delta x}{3\alpha- 2\alpha}= \frac{\Delta x}{\alpha}$ so (assuming the charge is uniformly distributed, the charge on it is $\displaystyle \frac{Q\Delta x}{\alpha}$. Putting those together, the force between P and Q is $\displaystyle k_e\frac{Q^2\Delta x\Delta\theta}{2\pi\alpha(x^2+ R^2)}$. The total force between ring and rod is the sum of those over all segments of the ring and rod. Taking the limit as $\displaystyle \Delta\theta$ and $\displaystyle \Delta x$ become infinitesimal, we get the double integral $\displaystyle \frac{k_eQ^2}{2\pi\alpha}\int_{x= \alpha}^{2\alpha}\int_{\theta= 0}^{2\pi}\frac{dxd\theta}{x^2+ R^2}$. topsquark likes this.

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