Recent content by tellmesomething

I see I think I get it a little ill have to keep rereading it to sort of absorb it fully....last question i promise, the arrow in your diagram in post #46 represents the perpendicular to the rotation axis..?

Ad you might've already guessed im very slow. I dont get you after "rotate by 120° about this cube diagonal". Your telling me that if I rotate the cube about its diagonal I would get a similar charge distribution yes...but shouldn't the net field still point towards the same direction...instead...

positive for all three..? Outward from a 2d page with a cuboid diagram... Like in the direction of the outward diagonal...

Oh I sort of just accepted the fact that clockwise means positive anti means negative etc...I never really questioned it..I get the reasoning now and how the above assumption could lead to wrong. Thankss..

I get the symmetry argument, we Can rename these axes and wed get the same magnitudes only different directions $$ E_{x} = \frac { \lambda} {4π \epsilon R√3} (- \hat I ) + \frac { \lambda} { 8π \epsilon R} (√3 + 1) (\hat j ) + \frac { \lambda} { 8π \epsilon R} (√3 + 1) (\hat k) $$ $$ E_{y} =...

It looks like I just messed up the limits..sorry for the waste of time I think im all good the correct results should be $$ E_{vert} = \frac { \lambda } { 4π \epsilon R√6} ( 1 + √3) $$ $$ E_{hori} = -\frac { \lambda} { 4 π \epsilon R√3} $$

The first one is correct I dont know why the latex reader missed the exponent... I didnt understand what exactly you wanted me to redo since your diagram was some what similar to mine... I get it now Phi is theta knot ... Since I didnt know how to type theta knot... Should have mentioned Sorry...

Thankyou!

This is what it should be showing incase it didnt get execute for you too..

Please let me know if you're seeing this executed ....im only seeing the tex commands at my end

Following this $$ R√2 tan \theta = x $$ $$ \frac {dx} {d\theta} = R√2 sec²\theta $$ $$ r= \frac { R√2} {cos \theta} $$ $$ dE = \frac { \lambda R √2 sec²\theta d\theta} { 4 π \epsilon ( \frac { R√2} {cos \theta} )^2 } $$ $$ \int dE cos \theta = \int_\phi^{(-π/2)} \frac {\lambda d\theta cos...

$$ E_{vert} =\frac { -\lambda } {4 π \epsilon R√6} (√3 + 1) $$ $$ E_{hori} =\frac {\lambda} { 4 π \epsilon R√3} $$