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1. Homework Statement
Griffiths' Introduction to Electrodynamics problem 2.10,
2. Homework Equations
Gauss' Law, ##\int_{S} \textbf{E}\cdot \textbf{dS} = \frac{Q_{\text{enc}}}{\epsilon_0}##
3. The Attempt at a Solution
It seems reasonable that the flux through the shaded surface and the front face of the cube are equal. I don't see any reason for them to be different. Assuming that they are equal, we should have that the flux through the front face added with the flux through the shaded face is ##\frac{q}{\epsilon_0}##, as the flux through the other faces is zero, since the electric field is tangential to those faces. Hence, we have ##\Phi_{\text{shaded}} = \frac{q}{2\epsilon_0}##. This is not the correct answer though! Why? Is it because the charge is on the boundary of the solid?
Let the charge as pictured be the origin, and say we solved this problem exactly for the charge at ##(\epsilon, \epsilon, \epsilon)## where ##\epsilon < \sqrt{3}##, then took the limit ##\epsilon \to 0##. Would we not get the flux to be ##\frac{q}{2\epsilon_0}##?
I understand the argument that we can consider a larger cube with q at the centre, and the final result comes to be ##\frac{q}{24 \epsilon_0}##, I would like to know why my method fails.
Griffiths' Introduction to Electrodynamics problem 2.10,
2. Homework Equations
Gauss' Law, ##\int_{S} \textbf{E}\cdot \textbf{dS} = \frac{Q_{\text{enc}}}{\epsilon_0}##
3. The Attempt at a Solution
It seems reasonable that the flux through the shaded surface and the front face of the cube are equal. I don't see any reason for them to be different. Assuming that they are equal, we should have that the flux through the front face added with the flux through the shaded face is ##\frac{q}{\epsilon_0}##, as the flux through the other faces is zero, since the electric field is tangential to those faces. Hence, we have ##\Phi_{\text{shaded}} = \frac{q}{2\epsilon_0}##. This is not the correct answer though! Why? Is it because the charge is on the boundary of the solid?
Let the charge as pictured be the origin, and say we solved this problem exactly for the charge at ##(\epsilon, \epsilon, \epsilon)## where ##\epsilon < \sqrt{3}##, then took the limit ##\epsilon \to 0##. Would we not get the flux to be ##\frac{q}{2\epsilon_0}##?
I understand the argument that we can consider a larger cube with q at the centre, and the final result comes to be ##\frac{q}{24 \epsilon_0}##, I would like to know why my method fails.
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