Okay I actually made an error like the comment by another user pinpointed, but I corrected this and get ##R^3_{SO}## to equal ##\sqrt {a^2 (2-2cos \theta)}^3 ## instead of 0.000181. However here is a picture showing an integral calculator online failing to solve this integral. Any ideas? This...
Okay I corrected this and get ##R^3_{SO}## to equal ##\sqrt {a^2 (2-2cos \theta)}^3 ## instead of 0.000181. However here is a picture showing an integral calculator online failing to solve this integral. Any ideas? This problem is really confusing me because considering I am working with a...
Note that the solution is 5625 V/m in z direction which is found easier using Gauss' law, but I want to find the same result using Coulombs law for confirmation.
Lets give the radius 0.04 the variable a = 0.04m.
##\rho## is the charge distribution distributed evenly on the surface of the...
Thus I assume that one slab has positive charge Q1
and the other slab has negative charge Q2 = -Q1
There are 4 cases for the electric field:
1. x <= -a
2. -a <= x <= 0
3. 0 <= x <= a
4. a <= x
The general case:
Charge Density ##\rho = \frac {Q} {V}##
Flux of E ##\phi_e = \oint \vec E \cdot d...
There is one cylinder. The charge Q is inside the cylinder, on the axis of the cylinder (is how I interpreted the question, and solved it). That is why I had Q = yh and not Q = y*(volume of cylinder). Because I assumed the charge is concentrated on the line in the cylinder
I begin by calculating the flux to be the flux of the cylinders lateral surface, which equals E*2*pi*p*h (p is the radius)
The other two surfaces have E ortogonal to dA, so their flux is 0.
Using Gauss law together with the calculated flux above, I get
Flux = Q/e
Flux = E*2*pi*p*h
Solve for E...