Question

A cube of sides length L contains a flat square plate also with sides of length L. The cube and square are in a Cartesian coordinate system. The square is placed at z=L/2 and extends from x=0 to x=L and from y=0 to y=L. The cube is placed such that it extends from x=0 to x=L, y=0 to y=L and z=0 to z=L. The flat square plate has a surface charge density that is given by -3xy (C/m2). Calculate the total electric flux passing through the sides of the cube.

Answer 1

Answer to Question #82041, Physics / Electromagnetism

A cube of sides length $L$ contains a flat square plate also with sides of length $L$. The cube and square are in a Cartesian coordinate system. The square is placed at $z = L/2$ and extends from $x = 0$ to $x = L$ and from $y = 0$ to $y = L$. The cube is placed such that it extends from $x = 0$ to $x = L$, $y = 0$ to $y = L$ and $z = 0$ to $z = L$. The flat square plate has a surface charge density that is given by $-3xy$ ($C/m2$). Calculate the total electric flux passing through the sides of the cube.

Solution.

By Gauss's law, the total electric flux passing through the sides of the cube:

$$\oint E dS = \frac{q}{\varepsilon_0}$$

where $q$ is the charge inside the cube.

So:

$$q = \int_{0}^{L} dx \int_{0}^{L} (-3xy) dy = -3 \int_{0}^{L} \left. \frac{xy^2}{2} \right|_{0}^{L} dx = - \frac{3L^2}{2} \cdot \left. \frac{x^2}{2} \right|_{0}^{L} = - \frac{3L^4}{4}$$

Answer:

$$\oint E dS = - \frac{3L^4}{4\varepsilon_0}$$

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