Gauss's Law long thin wire question

[Doc Al]

So would that use the equations I made earlier of

cos θ = x/r
sin θ = y/r

giving

E-x = E x/r

E-y = E y/r

TFM

Good. Plug in the value of E and you're done.

 

[TFM]

So:

E = lambda/(2*pi*r*epsilon-0)

Thus:

E-x = (lambda*x)/(2*pi*r^squared*epsilon-0)

and

E-y = (lambda*y)/(2*pi*r^squared*epsilon-0)

TFM

 

[borgwal]

finally...yes!

 

[TFM]

Thats good.

Thanks Everybody

TFM