Recent content by tbonepower07

thats how i had been doing it for a very long time, and it is correct but then I started trying to learn the more formal way to do line integrals and I couldn't understand why d \vec{s} was not equal to -dr, because it starts at infinity and goes to r. However, with frogpad's excellent help i...

Hey thanks for you help. I'll look into that book. The thing about these integrals is sometimes it looks like something you can sort of do mentally - like really simple applications of Gauss' Law - but it isn't - like the problem I just had.

okay, I think I have this figured out. The problem was that I didn't know how to do line integrals, because we were never taught them. I stared at a calculus book for a long time before I kind of figured it out. So - \int_C \vec{E} \cdot d \vec{s} = \int_a^b \vec{E(s)} \cdot \vec{s}'(t) dt...

thank you! whats a general rule for knowing what direction the differential vector should point? i.e, when calculating potential difference between two points for a given charge distribution?

yeah every single place i check says that i have the right formula V_B - V_A = - \int_a^b \vec{E} \cdot \vec{ds} I know this is the correct formula, however, I don't know how to deduce the correct answer when E = Q/(4 pi e_0 r^2) in the positive r direction, b is a point R, and a is at...

or how to compute electric potential using a line integral

the actual question was "where am i messing up"

hm, i thought that was supposed to when finding the potential, because infinity is where the potential is zero.

Okay, let me clarify that. V(r) = - \int_{\infty}^r \vec{E} \cdot d \vec{s} If S is the path from {\infty} to r, then d \vec{s} = - dr \hat{r} in polar coordinates. So \vec{E} \cdot d \vec{s} = E \hat{r} \cdot -dr \hat{r} = -E dr V(r) = - \int_{\infty}^r -E dr V(r) =...

I have become exceedingly confused by the various sign changes involved in computing the electric potential produced by a charge distribution, and I am sure I am simply forgetting a negative sign somewhere and am going crazy. However, V = -int from infty to r of E . ds ds is the...