I was attempting to get the E field of a disc with radius b--aka integrating from radius of 0 to b, but I don't think that's right. I'm confused by what you mean by "the general expression as a function of a and b." I don't understand how I can get a and b in the same equation...
From the E = k...
Ok so if I use E = k ∫(σ/b²) dA (from kq/r2 and Guass' Law using r=b) where dA=dθdb and θ is the angle between the point z and a point along radius a, I can differentiate it from a →b and from θ=0→360°. Is that what you mean?
Homework Statement
A washer made of nonconducting material lies in the x − y plane, with the center at the coordinate origin. The washer has an inner radius a and an outer radius b (so it looks like a disk of radius b with a concentric circular cut-out of radius a). The surface of the washer is...
Homework Statement
In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 260 kg and moves with speed v = 15 m/s. The loop-the-loop has a radius of R = 10 m. What is the minimum speed of the car so that it stays in contact with the track...
The forces acting on the top crate are the force of friction from the bottom crate, the force of gravity downward, and the normal force upward, correct? So the sum of the forces in the horizontal direction would be... only the force of friction?
Alright so I used Ffr=μm1g using μk=0.63 and got Ffr=(0.63)(20 kg)(9.8 m/s2) = 123.48 N
This would be the force of friction acting on the top crate, so would I then use Ffr=Ft-Fnet to find the Fnet, the sum of the forces acting on the top crate? I don't see another way to incorporate the new tension
Homework Statement
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 20 kg and the larger bottom crate has a mass of m2 = 80 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.79...