Another way to describe the difference between Coulomb's Law and Gauss' Law may be this (and please correct me if I'm wrong):
Coulomb's Law can be used to find the electric field at a particular point in space. However, where the charge distribution is more complex (like a cylinder) Coulomb's...
But if Gauss' law can be derived from Coulomb's Law (which only applies to point particles or charge distributions with spherical symmetry) how can it be applied to cylinders?
In Part C I did the same thing I did in Part A, just with cylinders. Part A is the general derivation of Gauss' Law from Coulomb's Law that I came across. It uses the area of the surface of a sphere. In Part C I did the same thing, just using the area of the surface a cylinder.
Doesn't EA =q/∈ (i.e. Gauss' law) come from somewhere? Of course, you can start with Gauss' Law and derive Coulomb's law from it (or something similar such as Part A Line 2 in the original post), but you can also start with Coulomb's Law and derive Gauss' law from it.
Are you saying that the way I derived Gauss' Law (using algebra) can only be done in very symmetric conditions, but for a more general derivation I should have used the integral form?
If I understand correctly, Gauss’ Law is (roughly) derived as follows:
Part A
Electric Flux = EA
E = q / (∈4πr^2)
A of the surface of a sphere is 4πr^2
They cancel out and therefore EA =q/∈
Line 4 seems to only apply to a sphere, as it is based on line 3.
Now, Gauss’ Law is applied to...
Here’s my calculation for #56:
1. Net torque is [(20kg)(9.8m/s^2) × (2.5m)] – [(5kg) (9.8m/s^2) × (5m)] = 490Nm -245Nm = 245Nm
2. The common acceleration is α = τ/I = 245Nm/[(20kg)(6.25m)] + (5kg)(25m)] = 245Nm/250kg m^2 = 0.98 rad/s^2
3. The COM is at 1m from the pivot. The...
I'm afraid telling isn't enough. You need to explain. I admit I was a bit stubborn, and I apologise for that, but those who explained to me why I was wrong and what my missteps were got further than those who just asserted I was wrong.
I also had a lot of other misunderstandings on the...
My original understanding was that τ=Iα should only apply where the torque and mass are at the same position - unless the moment of inertia changes. My mistake was that it was F=ma, not τ=Iα, which doesn't directly apply when the force and mass at different positions. You can repeat ad nauseam...
Ok here goes.
1. Torque on the left is 20 × 2.5 = 50. The torque on the right is 5 × 5 = 25. The net torque is 25.
2. α = τ/I. I = (20)(2.5)^2 + (5)(5)^2 = 250.
25/250 = 10 (the common angular acceleration)
3. The tangential acceleration is also 10 since the COM is at 1m from the pivot as...
Ok, let me try again. So, I have been rereading the whole thread and I think it’s all coming together. Hopefully I’m at the home stretch. Thanks for sticking through.
Suppose you have a 10m see-saw where the weight on the left is 10 kg and is 2.5m from the left end of the see-saw, and the...
Fair point. I thought the maths backed up my intuition (since the moment of inertia is derived from torque, a greater torque should result in a greater moment of inertia). My maths was wrong :(