# Calculating Electric Flux Φ w/o typical Gaussian surface

by chawk
Tags: electric flux, gauss law
 HW Helper P: 5,003 The fact that the electrostatic flux through any closed surface is equal to the charge enclosed divided by epsilon-naught can be easily proven using the divergence theorem and Coulomb's law (or Gauss' law in differential form). Start with Coulomb's law: $$\textbf{E}(\textbf{x})=\frac{1}{4\pi\epsilon_0}\int \rho(\textbf{x}')\frac{\textbf{x}-\textbf{x}'}{|\textbf{x}-\textbf{x}'|^3}d^3x'$$ Take the divergence of both sides of the equation, use the Leibniz integral rule, and the fact that $$\mathbf{\nabla}\cdot\frac{\textbf{x}-\textbf{x}'}{|\textbf{x}-\textbf{x}'|^3}=4\pi\delta(\textbf{x}-\textbf{x}')$$ and you find that $$\mathbf{\nabla}\cdot\textbf{E}(\textbbf{x})=\frac{1}{\epsilon_0}\int \rho(\textbf{x}')\delta(\textbf{x}-\textbf{x}')d^3x'=\frac{\rho(\textbf{x})}{\epsilon_0}$$ Which is Gauss' law in differential form (one of Maxwell's equations). From there, just apply the divergence theorem, along with the definition of charge density and you find \begin{aligned}\Phi &= \oint_{\mathcal{S}}\textbf{E}\cdot d\textbf{a} \\ &= \int_{\mathcal{V}}(\mathbf{\nabla}\cdot\textbf{E})d^3x \\ &= \frac{1}{\epsilon_0}\int_{\mathcal{V}}\rho(\textbf{x})d^3x \\ &= \frac{Q_{\text{enc}}}{\epsilon_0}\end{aligned} If this kind of vector calulus is too advanced for your current level, and you simply want to convince yourself with your cube example, just begin by expressing the electric field of the point charge Q at the origin in Cartesian coordinates by using the fact that the radial unit vector is $$\mathbf{\hat{r}}=\frac{x\mathbf{\hat{x}}+y\mathbf{\hat{y}}+z\mathbf{\ha t{z}}}{\sqrt{x^2+y^2+z^2}}$$ From there, just compute $\textbf{E}\cdot d\textbf{a}$ for one face of the cube and integrate over that face.
 P: 195 You want to calculate the flux through the sides of a cube of the field of a charge placed at its center. You were right while reasoning that the total flux through the cube would be 6 times the flux through a single side. So, you have got to calculate the flux through one side. Take, say, the top surface. Divide it into rectangular strips. Let x be constant for each strip. Let each strip be of thickness, dx. (the length of the strip = a is along y and the width = dx is along x). Choose one strip. Let it be at position 'x'. You'll have to further divide this strip (differential elements with sides dy & dx). Choose one differential element. Let it be at the point (x,y). Find out the value of E at this point. To make calculations simple choose the center of the square to be the origin. To calculate the flux, you'll need to find the normal component of E at this point. You should get $${(\frac{{a}^{2}}{4} + {x}^{2} + {y}^{2}})}^{\frac{3}{2}}$$ in the denominator. Integrate this w.r.t y to find the flux through the strip. Integrate this again w.r.t x to find the flux through the side.