Question

the total electric flux from a cubical box of side 13.0 cm is 2.75×10³ n·m²/c. part a: what charge is enclosed by the box? express your answer to three significant figures and include the appropriate units.

Explanation:

Step1: Recall Gauss's Law

Gauss's Law states that the total electric flux \(\Phi_E\) through a closed surface is equal to the enclosed charge \(Q_{enclosed}\) divided by the permittivity of free space \(\epsilon_0\), i.e., \(\Phi_E=\frac{Q_{enclosed}}{\epsilon_0}\). So we can rearrange this formula to solve for \(Q_{enclosed}\): \(Q_{enclosed}=\Phi_E\times\epsilon_0\).

Step2: Identify the values

We know that \(\Phi_E = 2.75\times 10^{3}\space N\cdot m^{2}/C\) and \(\epsilon_0=8.85\times 10^{- 12}\space C^{2}/(N\cdot m^{2})\).

Step3: Calculate the enclosed charge

Substitute the values into the formula: \(Q_{enclosed}=(2.75\times 10^{3})\times(8.85\times 10^{-12})\)
\[

$$\begin{align*} Q_{enclosed}&=2.75\times8.85\times 10^{3-12}\\ &=24.3375\times 10^{-9}\\ &=2.43375\times 10^{-8}\space C \end{align*}$$

\]
Rounding to three significant figures, we get \(Q_{enclosed} = 2.43\times 10^{-8}\space C\) (or \(24.3\space nC\)).

Answer:

The charge enclosed by the box is \(\boldsymbol{2.43\times 10^{-8}\space C}\) (or \(\boldsymbol{24.3\space nC}\)).