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Problem-Solving and Data Analysis Difficulty: Hard

For an electric field passing through a flat surface perpendicular to it, the electric flux of the electric field through the surface is the product of the electric field’s strength and the area of the surface. A certain flat surface consists of two adjacent squares, where the side length, in meters, of the larger square is $3$ times the side length, in meters, of the smaller square. An electric field with strength $29.00$ volts per meter passes uniformly through this surface, which is perpendicular to the electric field. If the total electric flux of the electric field through this surface is $4,640 volts dot meters$ , what is the electric flux, in $volts \cdot meters$ , of the electric field through the larger square?

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Explanation

The correct answer is $4,176$ . 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 $3 squared$ , 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 $9 x$ . Therefore, the flat surface of the two adjacent squares has a total area of $x plus 9 x$ , or $10 x$ . 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 $4,640 volts dot meters$ . 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 $29.00 left parenthesis 10 x right parenthesis equals 4,640$ , or $290 x equals 4,640$ , can be used to represent this situation. Dividing each side of this equation by $290$ yields $x equals 16$ . Substituting $16$ for $x$ in the expression for the area of the larger square, $9 x$ , yields $9 left parenthesis 16 right parenthesis$ , or $144$ , square meters. Since the area of the larger square is $144$ square meters, the electric flux, in $volts \cdot meters$ , 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 $left parenthesis 144 square meters right parenthesis left parenthesis StartFraction 29.00 volts Over meter EndFraction right parenthesis$ , or $4,176 volts dot meters$ .