Surface charge density of a plane

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The discussion focuses on calculating the surface charge density of a plane using Gauss's law. The enclosed charge is determined to be 9.6 x 10^-7 C, leading to a surface charge density of 3.04 x 10^-5 C/m² for a specified radius. Participants express difficulty in visualizing the system with multiple elements, specifically the interactions between a point charge, a plane, and a sphere. Clarifications are made regarding the positioning of the point charge relative to the sphere, impacting the calculations. The conclusion emphasizes that accurate diagrams can resolve many conceptual issues in such problems.
Guillem_dlc
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Homework Statement
A positive point charge of value ##q=+2\, \mu \textrm{C}## is located at coordinate point ##(0,0,5)\, \textrm{cm}##, above an infinite homogeneously charged plane located at ##z=0##. If the flux through a sphere of radius ##R=10\, \textrm{cm}## centered at the origin of coordinates is ##1,08\cdot 10^5\, \textrm{Nm}^2/\textrm{C}##, calculate the value of the surface charge density of the plane.
Relevant Equations
Gauss's Law
$$\phi_E=\dfrac{Q_{\textrm{enclosed}}}{\varepsilon_0}\Rightarrow Q_{\textrm{enclosed}}=9,6\cdot 10^{-7}\, \textrm{C}$$
$$Q_{\textrm{enclosed}}=\sigma S=\sigma \pi R^2\Rightarrow \sigma =\dfrac{Q_{\textrm{enclosed}}}{\pi (0,1^2)}=3,04\cdot 10^{-5}\, \textrm{C}/\textrm{m}^2$$

I have a lot of problems with the flux exercises. I have a hard time seeing how they act when I have more than one element in the system as in this case.
 
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Can you draw the point charge, the plane and the sphere?
 
Gordianus said:
Can you draw the point charge, the plane and the sphere?
1.png
 
Guillem_dlc said:
You have drawn the point charge outside of the sphere. Is this accurate according to the problem formulation?
 
Orodruin said:
You have drawn the point charge outside of the sphere. Is this accurate according to the problem formulation?
No, I see now. I think that I have the solution:

$$\phi_E = \dfrac{Q_{\textrm{enclosed}}}{\varepsilon_0} \Rightarrow Q_{\textrm{enclosed}}=9,6\cdot 10^{-7}\, \textrm{C}$$
$$\phi_{\textrm{enclosed}}=\sigma S=\sigma \pi R^2+q\Rightarrow \sigma=\dfrac{Q_{\textrm{enclosed}}-q}{S}=-33\, \mu \textrm{C}/\textrm{m}^2$$
 
A good drawing fixes many problems.
 
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