Video: Finding the Charge Contained within a Volume given the Electric Flux

The electric flux through a side of a cubical box is 8.8 Γ— 10⁴ Nβ‹…mΒ²/C. What is the total charge enclosed by the box?

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Video Transcript

The electric flux through a side of a cubical box is 8.8 times 10 to the fourth newtons meter squared per coulomb. What is the total charge enclosed by the box?

We can call the electric flux given, 8.8 times 10 to the fourth newtons meter squared per coulomb, 𝛷 sub 𝐄. We want to solve for the total charge that’s enclosed by the box. We’ll call that charge 𝑄.

If we draw a sketch of our cubical box, we can imagine that there is both positive and negative charges inside. And we just know that the total charge is represented by capital 𝑄. If we recall Gauss’s law, this law tells us that the total electric flux over a closed surface, 𝛷 sub 𝐄, is equal to the charge enclosed by that surface over πœ€ naught, the permittivity of free space, whose value we can take to be equal to 8.85 times 10 to the negative 12 farads per meter.

This electric flux shown in Gauss’s law is equal to the total flux through an entire closed surface. And remember, in our case, 𝛷 sub 𝐄 is the flux only through one side of our cubical box. So if we want to solve for the total net charge enclosed in our cube, then we would need to take our value for 𝛷 sub 𝐄 and multiply it by six in order for it to be equal to that total charge 𝑄 over πœ€ naught.

If we rearrange this equation to solve for 𝑄, we see it’s equal to six times 𝛷 sub 𝐄 times πœ€ naught. When we plug in for 𝛷 sub 𝐄 and πœ€ naught and enter these values on our calculator, we find, for a value of 𝑄, 4.7 times 10 to the negative sixth coulombs. That’s the total charge enclosed in the cubicle box.

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