Still more problems with Gauss Law

AI Thread Summary
The discussion revolves around applying Gauss's Law to find the electric field at two points related to a spherical cavity with a charge at its center. For point P1, the calculations align with Coulomb's law, allowing for straightforward application of Gauss's Law. However, for point P2, there is uncertainty about using a spherical Gaussian surface, as the enclosed charge must be determined first. The conversation highlights the importance of considering induced charges on the inner surface of the metal when calculating the electric field in such scenarios. Understanding these concepts is crucial for correctly applying Gauss's Law in electrostatics.
stunner5000pt
Messages
1,443
Reaction score
4
Ok I'm realllly unsure about this!

If a spherical cavity of radius 3.66cm in a piece of metal (kind of cube, but not a perfect cube) has a charge of +Q at it's centre and there is a point P1 located half way between the spherical cavity and it's surface and a point P2 located in the metal piece , use gauss Law to find the electric field at
a) Point P1
b) point P2

a) This is fine for me

Let e0 = permittivity of free space epsilon 0

e0 SurfaceIntegral (E dA) = q
e0 E (4 pi r^2 ) = q
E = kq /r^2
which is coulomb's law derived! and just plug and away i go!

b) But for point P2

err can i actually use a spherical surface here??

as in do the same thing as before but use a bigger value for r i.e. radius of sphere + distance of sphere to point P2

appropriate? or not?
 
Physics news on Phys.org
Umm...the answer to that would be : NOT YET. Mostly because you don't yet know what charges are enclosed by this Gaussian surface.

Read what your text has to say about induced charges and electric fields inside a piece of metal.
 
You can still use a spherical surface though.

Claude.
 
Claude Bile said:
You can still use a spherical surface though.

Claude.


i'm not quite sure how i would use it , so the shere would have a radius of 3.66cm + A ??

e0 E 4 pi (3.66+A)^2 = 126 x 10^-9

something liek this??
 
You need to take into account the induced charges on the inner surface of the metal.

Claude.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Back
Top