# Charge enclosed Guassian surface

by craka
Tags: charge, enclosed, guassian, surface
 P: 20 1. The problem statement, all variables and given/known data Question states "Total electric flux from a cubical box 0.28m on a side is 1.84x10^3 Nm^2/C. What is the charge enclosed by the box 2. Relevant equations Gauss's Law $$\begin{array}{l} \phi _E = \vec E \bullet \vec A \\ \oint {\vec E \bullet d\vec A = \frac{{Q_{enclosed} }}{{\varepsilon _0 }}} \\$$ 3. The attempt at a solution $$\begin{array}{l} Q_{enclosed} = (4\pi r^2 )\varepsilon _0 \phi _E \\ \varepsilon _0 = 8.85 \times 10^{ - 12} \\ \phi _E = 1.84 \times 10^3 \\ r = 1.2 \times 0.28 = 0.14 \\ Q_{enclosed} = 4.01 \times 10^{ - 9} Nm^2 /C \\ \end{array}$$ But the answer is given as 1.63*10^-8 Nm^2/C What I have I done wrong here?
 HW Helper P: 2,688 I noticed the following mistakes at first glance: 1) You have a cubical box, yet you are using the surface area of a sphere: $4\pi r^2$ You need to use the surface area of this cube, which is the sum of the area of each side. 2)The correct answer you give has the wrong units. The units should be C, not Nm^2/C. Check that answer to make sure it corresponds to this problem. Are either of these the problem?