Recent content by holdmyhand

I found it! Infinite thanks for your help and patience!

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)?

The maximum θ is 45° and the minimum is -45°, no?

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...

What do you mean by this?

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...