Video: Determining the Charge Density of a Large Sheet from the Net Electric Flux Produced through a Square Area Parallel to It

The electric flux through a square-shaped area of side 5.0 cm near a large charged sheet is found to be 3.00 × 10⁻⁵ N⋅m²/C when the area is parallel to the plate. Find the charge density on the sheet.

03:26

Video Transcript

The electric flux through a square-shaped area of side 5.0 centimeters near a large charged sheet is found to be 3.00 times 10 to the negative fifth newton meters squared per coulomb when the area is parallel to the plate. Find the charge density on the sheet.

We can call the length of the sides of the square, 5.0 centimeters, 𝐿. And we can call the electric flux through the square, 3.00 times 10 to the negative fifth newton meters squared per coulomb, 𝜙 sub 𝐸.

We want to solve for the charge density on the sheet, what we’ll call 𝜎. We can begin by drawing a diagram. We’re told we have a large charged plate. And we’ve assigned the charge density on the plate to be 𝜎. The plate extends very far in all directions. Essentially, it’s an infinite charged plate.

Very nearby and parallel to the charged plate, we have a square area of side length 𝐿, given as 5.0 centimeters. The charge on the plate creates electric field lines that move away from the plate and through the area of the square. We’re told the flux through that square and want to work backwards to solve for 𝜎.

To do that, we can recall Gauss’s law which says that the electric flux through an enclosed surface equals the charge that surface encloses divided by 𝜖 naught, the permittivity of free space. In our scenario with our charged sheet, it’s important to realize that the charge on the sheet sends electric field lines out in both directions perpendicular to the plane of the sheet.

This is important because when we consider what to plug in for 𝑄, instead of plugging in 𝜎 that we want to solve for times the area of our square, we’ll want to use only half that value because our square is only on one half of the charged plate. The electric field lines pointing in the other way from the charge plate don’t pass through the square at all and therefore don’t contribute to the flux.

Substituting 𝜎𝐴 over two in for 𝑄, the electric flux, 𝜙 sub 𝐸, corresponding to our measured value given in the problem statement equals 𝜎 times 𝐴 where 𝐴 is 𝐿 squared divided by two 𝜖 naught, where 𝜖 naught, a constant, we’ll treat as exactly 8.85 times 10 to the negative 12 farads per meter.

We now want to rearrange this equation to solve for 𝜎. It’s equal to two 𝜖 naught times 𝜙 sub 𝐸, the flux, over 𝐿 squared. 𝜖 naught is a known constant. 𝜙 sub 𝐸 is given in the problem statement, as is 𝐿.

We’re now ready to plug in and solve for 𝜎. When we do. we’re careful to use units of meters for the length of the side of our square 𝐿. Plugging these values into our calculator, we find that 𝜎, to two significant figures, is 2.1 times 10 to the negative 13th coulombs per meter squared. That’s the charge density of this extended sheet.

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