Recent content by an_single_egg

Thank you!

Flux = BA but BA is technically the dot product... so yes, but when the magnetic field is NOT perpendicular you need to multiply BA by ##cos\theta##, where theta is the angle between B and A. So you could also changed induced current by tilting the loop?

Homework Statement My class was watching a video on electricity and magnetism, and after showing how current is induced in a loop of wire when a magnet moves through it, our teacher asked us if there is another way to induce current in a loop of wire - he told us that there were in fact two...

Why am I thinking about infinite distances for part b? If ##r_0## is a finite radius and r<##r_0##, isn't everything contained within the sphere? The potential for any ##r## > ##r_0## is ## \frac {KQ} {r^2}##.

Ok, I think I've got it: ## E = \frac {Qr^3} {4\pi \epsilon_0 r_0^5} ## ## V_{ba} = \int_r^{r_0} \frac {Qr^3} {4\pi \epsilon_0 r_0^5}## ##= \frac {Q} {16\pi \epsilon_0 r_0^5} (r_0^4 - r^4)## ##= \frac {Q} {16 \pi \epsilon_0 r_0} (1 - \frac {r^4} {r_0^4})## Now... I know that 1 should be a 5...

I'm trying to plug into Gauss's Law, $$ E \int dA = \frac {Q_{enc}} {\epsilon_0} $$ $$ E \cdot (4\pi r^2) = \frac {4k \pi r^5} {5\epsilon_0} $$ And then solving for ##E,## I tried to find ##V_{ba}## by integrating E from ##r \rightarrow r_0##, but that's not correct. I think this is confusing...

My book is telling me that the answer for part b is $$ V = \frac {Q} {16 \pi \epsilon_0 r_0} (5 - \frac {r^4} {r_0^4})$$ If I plug in ##\frac {5Q} {4 \pi r_0^5}## for k, I'm not left with anything to give me that answer... I am definitely missing something but don't know what. :(

Just solving for k, $$k = \frac {5Q} {4\pi r^5}$$ Or did you mean something in my integral wasn't right? I'm a little confused how to get k in terms of ##Q## and ##r_0## , if the integral is from ##0## to ##r##. Do I need to be taking it from ##r## to ##r_0## ? I'm really sorry, I'm just...

Sorry, I guess I didn't fully understand what that portion of the problem meant. I also realized I'm supposed to be solving in terms of Q, r, and r_0 and constants, so I'll be trying to solve it correctly now, oops! Does this still apply? If I use that integration, then $$ Q = \int_0^r kr^2 dV...

Ok, so: Q = (0→r)∫ρEdV =4/3ρEπr3 Then, E⋅∫dA = Qenc/ε0 E(4πr2)=(4ρEπr3)/(3ε0) E=(ρEr3)/(3r2ε0) E=(ρEr)/3ε0 And then, Vba = -(r→r0)∫(ρEr)dr/3ε0 Vba = -ρE/6ε0 * (r02-r2) Is that right?

Homework Statement A nonconducting sphere of radius r0 carries a total charge Q. The charge density ρE increases as the square of the distance from the center of the sphere, and ρE=0 at the center. a) Determine the electric potential as a function of the distance r from the center of the...