Three 20 cm-long 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?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Three 20 cm-long 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?

Physics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

All i need is help on how to start it.
Well 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.
\[\oint\vec E\cdot d\vec A=\frac q{\epsilon_0}\]\[E(2\pi r\ell)=\frac{\lambda\ell}{\epsilon_0}\to\vec E=\frac{\lambda}{2\pi r}\vec r\]so now that we have our field formula we can get our three vectors at the point and superimpose them.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

by symmetry the situation will be this way:|dw:1329356841736:dw|
|dw:1329356974346:dw|the angles should be 60/2=30degrees
ok, im not sure if i know that equation yet. so how do you use it?
don't know Gauss' law? I wonder how else it can be solved...
The 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\[\sum\vec E=\vec E_1+\vec E_2+\vec E_3=|E|\cos\theta_1+|E|\cos\theta_2+|E|\cos\theta_3\]\[=|E|(\cos\theta_1+\cos\theta_2+\cos\theta_3)=\frac{\lambda}{2\pi r}[\cos(30)+\cos(0)+\cos(-30)]\]
lambda is the charge per unit length by the way...
ok i understand this so far. Thanks
welcome, let me know if you get stuck
Real Quick, how do i find the radius in this situation?
the 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
got it
I hate to say it but i did everything and i still got it wrong. In the end i got 136.1.
let me give it a try...
sorry, I get\[1.76\times10^{-8}\frac Vm\]
nope still wrong. :/
ok, closer investigation it seems strange to me that you got such a high number though, there isn't than much charge...
I have no clue.
ok I'm really hoping it's about\[3.51\times10^{-7}\]
wait no!
I forgot e0 all the way through!
sadly nope. i only have one try left. I'm not sure whats happening
Oh I didn't realize you had limited tries, sorry :(
ok 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
ok sounds good. And thank you soooooo much for doing that
(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...
Well, the correct answer was 3.8*10^4. i didnt get it right but i just thought i would let you know
Thanks for your help anyways
the above comes to 3.97*10^4 ...
sorry, but I stand by the formulation, and would have to use the same techniques to solve your latest question above
sucks I forgot epsilon0 for so long though, otherwise might have figured it out :/
I'll ask me teacher when i go to school

Not the answer you are looking for?

Search for more explanations.

Ask your own question