For an electric field passing through a flat surface perpendicular to it, the electric flux of

The correct answer is $\mathrm{4,176}$. It’s given that the side length of the larger square is $3$ times the side length of the smaller square. This means that the area of the larger square is $3^2$, or $9$, times the area of the smaller square. If the area of the smaller square is represented by $x$, then the area of the larger square can be represented by $9x$. Therefore, the flat surface of the two adjacent squares has a total area of $x+9x$, or $10x$. It’s given that an electric field with strength $29.00$ volts per meter passes uniformly through this surface and the total electric flux of the electric field through this surface is $\mathrm{4,640} \text{volts}·\text{meters}$. Since it's given that the electric flux is the product of the electric field’s strength and the area of the surface, the equation $29.00\left(10x\right)=\mathrm{4,640}$, or $290x=\mathrm{4,640}$, can be used to represent this situation. Dividing each side of this equation by $290$ yields $x=16$. Substituting $16$ for $x$ in the expression for the area of the larger square, $9x$, yields $9\left(16\right)$, or $144$, square meters. Since the area of the larger square is $144$ square meters, the electric flux, in $\text{volts}·\text{meters}$, of the electric field through the larger square can be determined by multiplying the area of the larger square by the strength of the electric field. Thus, the electric flux is $\left(144 \text{square} \text{meters}\right)\left(\frac{29.00 \text{volts}}{\text{meter}}\right)$, or $\mathrm{4,176} \text{volts}·\text{meters}$.