So from that point onwards, I have kλ/r * (sin45 - sin(-45)), but I'm still lacking λ. How do I calculate the value of lambda? Would I just divide Q by the length of the semicircle (2πr/4)?
I could simply try integrating ∫dq because λ,k and r are all constant, but I don't know the bounds in that case, which makes it hard to carry on.
I have to go to another class right now, so I'll be back in ~2 hrs. I really appreciate your help, though!
Well, the field wouldn't go from full strength to zero, I imagine. 25μC is uniformly distributed across the rod, so of course EsinΘ would end up by cancelling out. I essentially just need to find the sum of the electric field throughout the rod and I understand that, but I'm having trouble...
Well, the quarter circle extends over an angle of 90 degrees, so I thought going from 0 to pi/2 would simplify it.
I just tried integrating from (3pi/8 to 5pi/8) and the answer to the integral is 0, so that definitely doesn't help.
The total charge of the quarter circle is 25μC.
So I just did
Q = λ ∫cosΘdl from 0 to pi/2
Q= λ ∫r*cosΘdΘ from 0 to pi/2
Q= λr * (sin(pi/2) - sin(0)
Q= λ = 25μC/m
which means...
(9*10^9) / 1^2 * 25μC/m = 2.25*/10^11
still the wrong answer.
Homework Statement
So a quarter circle has a total charge of 25μC and has a 1m radius.
I need to find the resultant electric field at point P.
Homework Equations
The Y components will cancel here due to symmetry so we are left only caring about the x components.
dE= (kdq)/(r^2) rˆ
dq= λdl...