2.10 A charge \(q\) sits at the back corner of a cube of side length \(a\), as shown in the figure. What is the flux of \(\vec{E}\) through the shaded side?

The electric field at some point on the shaded side is

$$\vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{q}{ ( a^2 + x^2 + z^2 )^{3/2} } \left( x \, \hat{i} + a \, \hat{j} + z \, \hat{k} \right)$$

$$\vec{E} \cdot \hat{j} = \frac{ a q }{4 \pi \epsilon_0} \frac{1}{ ( a^2 + x^2 + z^2 )^{3/2} }$$

$$\Phi = \int_S \vec{E} \cdot d\vec{A} = \int_0^a \int_0^a \vec{E} \cdot \hat{j} \, dx \, dz = \frac{ a q }{4 \pi \epsilon_0} \int_0^a \int_0^a \frac{1}{ ( a^2 + x^2 + z^2 )^{3/2} } \, dx \, dz$$

This form is very similar to a form encountered in problem 2.4 with the square loop. Again, using a trigonometric substitution

$$x = \sqrt{a^2 + z^2} \tan \theta \hspace{10mm} dx = \sqrt{a^2 + z^2} \sec^2 \theta \, d \theta$$

$$\Phi = \frac{ a q }{4 \pi \epsilon_0} \int_0^a \left[ \int \frac{ \sqrt{a^2 + z^2} \sec^2 \theta \, d \theta}{ (a^2 + z^2)^{3/2} ( 1 + \tan^2 \theta)^{3/2}} \right] \, dz$$

$$\Phi = \frac{ a q }{4 \pi \epsilon_0} \int_0^a \left[ \int \frac{ \sec^2 \theta \, d \theta}{ (a^2 + z^2) \sec^2 \theta} \right] \, dz = \frac{ a q }{4 \pi \epsilon_0} \int_0^a \left[ \int \frac{ \cos \theta \, d \theta}{ (a^2 + z^2) } \right] \, dz$$

$$\Phi = \frac{ a q }{4 \pi \epsilon_0} \int_0^a \left[ \int \frac{ \cos \theta \, d \theta}{ (a^2 + z^2) } \right] \, dz = \frac{ a q }{4 \pi \epsilon_0} \int_0^a \left[ \frac{ \sin \theta }{ (a^2 + z^2) } \right] \, dz$$

$$\Phi = \frac{ a q }{4 \pi \epsilon_0} \int_0^a \left[ \frac{ x }{ (a^2 + z^2) \sqrt{ a^2 + x^2 + z^2 }} \Big|_0^a \right] \, dz = \frac{ a q }{4 \pi \epsilon_0} \int_0^a \frac{ a }{ (a^2 + z^2) \sqrt{ 2 a^2 + z^2 }} \, dz$$

Using Mathematica

$$\Phi = \frac{ a q }{4 \pi \epsilon_0} \frac{ \pi }{ 6 a } = \frac{ q }{ 24 \epsilon_0 }$$