Electric Flux Problem: Can't Calculate Answer - Please Help!

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To calculate the electric flux through one face of a cube with a point charge at its center, first determine the total flux through the entire surface using Gauss's law, which states that the total flux is equal to the charge divided by the permittivity of free space (Φ = q/ε₀). For a charge of 9.6X10^-6 C, the total flux through the cube is approximately 1.81X10^5 Nm²/C. Since the cube has six identical faces, the flux through one face is the total flux divided by six, leading to a flux of about 3.02X10^4 Nm²/C per face. The symmetry of the cube simplifies the calculation, confirming that the approach is valid despite the assumption about the direction of E and dA. Understanding these principles allows for an accurate determination of the electric flux.
crazynut52
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I've been thinking about this problem for a while now, and I can't get it to work out. A 9.6X10^-6 C point charge is at the center of a cube with sides of length .5m What is the electric Flux through one of the six faces of the cube? I know the answer is 1.81X10^5 but I can't figure out how to get there. Please Help!
 
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Hint 1: What would the total flux be through the entire cubic surface?

Hint 2: Invoke symmetry!
 
\oint {E \cdot dA} = \frac{q}{\epsilon_0}

Let's say B = the area of one side then 6B = A if A is the total surface area so...

\oint {E \cdot da} = \frac{q}{\epsilon_0} = 6b \times E

the flux through B is therefore what? (I Realize this isn't entirely correct because I am assuming E and dA are parallel but the answer remains the same through symmetry)
 
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sure is easy when you figure it out... thanks for the help
 

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