Ya, what I said first was right. I just got a little confused because they gave extra info... I hate when that happens.
"I know one thing - that I know no thing." - lol, paradox
Homework Statement
A test charge +q is brought to a point A a distance r from the center of a sphere having a net charge +Q. Next, a test charge +2q is brought to a point B a distance 2r from the center of the sphere.
Compared with the electrostatic potential at A, the electrostatic potential...
Now I've ran into a new problem in the same set, here's the question:
What is the surface charge density inside the hollow cylinder? Answer in units of nC/m2
Because the field on the inside of the hollow cylinder is zero, the net charge must be zero (since the area is definitely not zero)...
They round to 6 sig digs & they accept any answer within a 1% error. You can change the answer by one and it'll still be within that 1% error range. Also they don't round until the end which is what I always do so I should get their exact answer. My guess is that the answer is either negative or...
The answer is NOT 4824.9 N/C. I haven't tried the rest, I'm waiting until I get this one correct
Here's where I looked to see if my method of solving it was correct:
http://www.physics.wisc.edu/undergrads/courses/spring09/248/HWSolutions/HW6Solutions.pdf
Short of scanning the image for the picture, that's it. I'm thinking it might be negative but I wanted to see if anyone could catch an error in what I did
Not if you're inside, but the radius I'm given is 1.9 cm and the diameter is 3.6 cm (1.8 cm radius). Therefore, the point it's asking for is outside the inner cylinder
Using a cylinder as the Gaussian surface, you get:
E*2\pi rl = Q/\epsilon_{0}
and Q/l=\lambda
therefore, Q=l\lambda
substitute and divide both sides by 2\pi rl:
E=l\lambda/2\pi rl\epsilon_0
cancel out the ls:
E=\lambda/2\pi r\epsilon_0
Homework Statement
The figure shows a portion of an infinitely long, concentric cable in cross section. The inner conductor carries a charge of 5.1 nC/m and the outer conductor is uncharged.
What is the electric field 1.9 cm from the central axis? The permittivity of free space is 8.85419e-12...
The Gaussian surface is a plane (flat plane) and I actually just found the powerpoint from my teacher and there was a slide on this. I understand what you need to do to solve the problem, I just don't really understand why Coulomb's law does not apply.
I'm guessing you are talking about Gauss's Law, but if you integrate, you get the integral of 2ko<pi>dr/r2. Therefore the force does vary by difference, that's Coulomb's Law
I'm not saying you're wrong, but I'm just finding things conflicting with each other
You are assuming the wire is a point charge, which it is not. You would have to integrate:
2 (Integral from 0 to infinity) kdq/r^2
I hope you know what to do from there