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anonymous
 4 years ago
Three 20 cmlong rods form an equilateral triangle. Two of the rods are charged to + 14 nC, the third to  14nC.What is the electric field strength at the center of the triangle?
anonymous
 4 years ago
Three 20 cmlong rods form an equilateral triangle. Two of the rods are charged to + 14 nC, the third to  14nC.What is the electric field strength at the center of the triangle?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0All i need is help on how to start it.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1Well it seems we're gonna want to start with Gauss' law to find the field due to each rod. I'm assuming that is applicable here, but that is a simplification.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1\[\oint\vec E\cdot d\vec A=\frac q{\epsilon_0}\]\[E(2\pi r\ell)=\frac{\lambda\ell}{\epsilon_0}\to\vec E=\frac{\lambda}{2\pi r}\vec r\]so now that we have our field formula we can get our three vectors at the point and superimpose them.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1by symmetry the situation will be this way:dw:1329356841736:dw

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1329356974346:dwthe angles should be 60/2=30degrees

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, im not sure if i know that equation yet. so how do you use it?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1don't know Gauss' law? I wonder how else it can be solved...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1The way it would go from here is that we have the formula for the electric field a distance r from the rods use geometry to find this point, then add the three vectorially: define straight down as zero: First note that the point is equidistant from each side of the triangle, so the magnitude of the fields there will be the same for each rod. Also note that by symmetry the horizontal components cancel, and the vertical components sum will be\[\sum\vec E=\vec E_1+\vec E_2+\vec E_3=E\cos\theta_1+E\cos\theta_2+E\cos\theta_3\]\[=E(\cos\theta_1+\cos\theta_2+\cos\theta_3)=\frac{\lambda}{2\pi r}[\cos(30)+\cos(0)+\cos(30)]\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1lambda is the charge per unit length by the way...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok i understand this so far. Thanks

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1welcome, let me know if you get stuck

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Real Quick, how do i find the radius in this situation?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1the distance from each rod to the center is found through geometry if you forgot how, break the triangle in half and use the pythagorean theorem to find the height. then divide the height by 2 that will be the midpoint of an equilateral triangle

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I hate to say it but i did everything and i still got it wrong. In the end i got 136.1.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1let me give it a try...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1sorry, I get\[1.76\times10^{8}\frac Vm\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1ok, closer investigation it seems strange to me that you got such a high number though, there isn't than much charge...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1ok I'm really hoping it's about\[3.51\times10^{7}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I forgot e0 all the way through!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sadly nope. i only have one try left. I'm not sure whats happening

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1Oh I didn't realize you had limited tries, sorry :(

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1ok I will just write out all my work and you can look for a mistake... I stand by my formula, but I omitted epsilon zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok sounds good. And thank you soooooo much for doing that

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1(everything in meters) linear charge density:\[\lambda=\frac{14\times10^{9}}{0.2}=7\times10^{8}\]finding the distance to the midpoint r:\[a^2+b^2=c^2\to0.2^2=0.1^2+a^2\]\[ a=\sqrt{0.03}=0.1\sqrt3\]\[r=a/2=0.05\sqrt3\]and then all that business I derived earlier\[E=\frac{\lambda}{2\pi\epsilon_0r}[\cos(30)+\cos0+\cos(30)]\]\[=\frac{7\times10^{8}}{2\pi(8.85\times10^{12})(0.05)\sqrt3}(\sqrt3+1)\]let me know if you see something strange...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, the correct answer was 3.8*10^4. i didnt get it right but i just thought i would let you know

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks for your help anyways

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1the above comes to 3.97*10^4 ...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1sorry, but I stand by the formulation, and would have to use the same techniques to solve your latest question above

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1sucks I forgot epsilon0 for so long though, otherwise might have figured it out :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'll ask me teacher when i go to school
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