Gauss's Law - Closed Half Cylinder

A particle with charge q = + 10 q = +10 is at position ( x , y , z ) = ( 1 2 , 1 3 , 2 3 ) (x,y,z) = \Big(\frac{1}{2},\frac{1}{3},\frac{2}{3} \Big) . A closed surface consists of four sub-surfaces put together.

The first sub-surface is a half-disk in the plane z = 0 z = 0 :

x 2 + y 2 1 x 0 z = 0 x^2 + y^2 \leq 1 \\ x \geq 0 \\ z = 0

The second sub-surface is a half-disk in the plane z = 1 z = 1 :

x 2 + y 2 1 x 0 z = 1 x^2 + y^2 \leq 1 \\ x \geq 0 \\ z = 1

The third sub-surface is a half cylinder:

x 2 + y 2 = 1 x 0 0 z 1 x^2 + y^2 = 1 \\ x \geq 0 \\ 0 \leq z \leq 1

The fourth sub-surface is a rectangle:

x = 0 1 y 1 0 z 1 x = 0 \\ -1 \leq y \leq 1\\ 0 \leq z \leq 1

Let the electric fluxes through the four sub-surfaces be ϕ 1 , ϕ 2 , ϕ 3 , ϕ 4 \phi_1,\phi_2,\phi_3,\phi_4 . Determine the following ratio:

ϕ 1 ϕ 2 ϕ 3 ϕ 4 ϕ 1 + ϕ 2 + ϕ 3 + ϕ 4 \frac{\phi_1 \, \phi_2 \, \phi_3 \, \phi_4}{ \phi_1 + \phi_2 + \phi_3 + \phi_4}

Details and Assumptions:
1) Electric permittivity ϵ 0 = 1 \epsilon_0 = 1
2) Use outward-facing normal vectors


The answer is 2.9.

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1 solution

I will not elaborate all the things that how I reached these integral expression, if you see old questions of this type only then @Karan Chatrath Sir has explained it in a very explanatory manner that how to reach this integral expressions. For calculating ϕ 1 \phi_{1} and ϕ 2 \phi_{2} I have used polar coordinates and for ϕ 4 \phi_{4} I have used simple cartesian coordinates.

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