The electric field in a region is given by $\overrightarrow{\mathrm{E}}=(2 \hat{i}+4 \hat{j}+6 \hat{k}) \times 10^3 \mathrm{~N} / \mathrm{C}$. The flux of the field through a rectangular surface parallel to $x-z$ plane is $6.0 \,\mathrm{Nm}^2 \mathrm{C}^{-1}$. The area of the surface is _____________ $\mathrm{cm}^2$.

<p>To calculate the area of the surface through which the electric flux is determined, we use the given electric field and the specified surface orientation. The electric field is expressed as $\overrightarrow{\mathrm{E}} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^3 \, \mathrm{N/C}$.</p> <p>Since the surface is parallel to the $x-z$ plane, its area vector $\overrightarrow{\mathrm{A}}$ is directed along the $y$-axis, making it $\mathrm{~A} \hat{\mathrm{j}}$.</p> <p>The electric flux $\phi$ through the surface is given by:</p> <p>$ \phi = \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{~A}} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^3 \cdot \mathrm{~A} \hat{\mathrm{j}} $</p> <p>Calculating the dot product, the only component contributing to the flux is the $y$-component:</p> <p>$ \phi = (4 \times 10^3) \mathrm{~A} $</p> <p>Given that the electric flux $\phi$ is $6.0 \, \mathrm{Nm}^2/\mathrm{C}$, we can equate and solve for $\mathrm{A}$:</p> <p>$ 6 = 4 \times 10^3 \mathrm{~A} $</p> <p>$ \mathrm{A} = \frac{6}{4 \times 10^3} = 1.5 \times 10^{-3} \, \mathrm{m}^2 $</p> <p>Converting this area from square meters to square centimeters:</p> <p>$ \mathrm{A} = 1.5 \times 10^{-3} \, \mathrm{m}^2 = 15 \, \mathrm{cm}^2 $</p>