# Nonuniformly charged semicircle

1. Jan 24, 2010

### pittmed80

A non-uniformly charged semicircle of radius R=18.5 cm lies in the xy plane, centered at the origin, as shown. The charge density varies as the angle θ (in radians) according to λ = − 4.42 θ, where λ has units of μC/m.What is the total charge on the semicircle?

I calculate Q to be (lambda)*L, or Q=(-4.42*(theta))*(.185*(theta)). I then ran this over the integral from theta=zero to theta=(pi). I got an answer in micro-Coulombs so I converted it to Coulombs to answer. I ended up with -8.4513E10^-6, but it says that It's incorrect...help!?!?! due at 11 tonight!

2. Jan 24, 2010

### rgk13

You know that a very small section of the charge would be dQ. You also know that the charge Q = (lamda)*s (this was wrong under your relevant equations (the arc length of that segment)). So... dQ = (lamda)*ds using some basic trig and what not you know that ds = r*d(theta) so then you get...

dQ = lamda*R*dTHETA and you're give your value of lambda so sub that in and you get...

dQ = -4.42(Theta)*R*d(theta) so integrate with respect to theta from 0 to pi b/c it's a semi-circle so you get...

Q = -4.42*R[ integral from 0 to pit of (theta d theta) ]

or

q = -4.42*R*(theta)^2/2 from 0 to pi...

Hope this helps :)