- #1
CAF123
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Homework Statement
A point charge of negative polarity is located at the centre of a cubic Gaussian surface with edges of length ##0.5m##.
Calculate the electric flux through one of the faces of the surface.
What would happen if the charge was moved 10cm to the right?
Homework Equations
Gauss' Law for closed surfaces
The Attempt at a Solution
From intuition, I can see that the answer is simply ##\frac{-q}{6\epsilon_o}##. However, I want to show this.
Denote the length of a side by ##a##. The E-field lines end within the closed surface, so the net flux should be negative (as I get above). Through one face, I have ##\Phi = -EA##, where ##A## is the area of a face and the negative is there because ##E## and ##A## are antiparallel. The E field of the charge a distance a/2 away is given by $$\frac{-q}{4\pi \epsilon_o (a/2)^2}$$ and so I get after some cancellation:$$\Phi = \frac{q}{\pi \epsilon_o}$$ which is not quite right. (I have a π still and the flux is positive). Where did I go wrong?
For the second question, the situation would no longer be symmetric. I am not entirely sure what the question wants me to do here though. Since the cube is a closed surface, I feel the net flux would still be simply ##-q/\epsilon_o##, although I am not sure about the individual faces.
Many thanks