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 - meters , what is the electric flux, in volts - meters , of the electric field through the larger square? __

To find the electric flux through the larger square, we need to use the relationship between the total electric flux and the areas of the squares.<br /><br />Let's denote the side length of the smaller square as \( s \). Therefore, the side length of the larger square is \( 3s \).<br /><br />The area of the smaller square is \( s^2 \), and the area of the larger square is \( (3s)^2 = 9s^2 \).<br /><br />The total electric flux through the surface is given by:<br /><br />\[ \text{Total Electric Flux} = \text{Electric Field Strength} \times \text{Total Area} \]<br /><br />Given:<br />- Electric Field Strength = 29.00 volts/meter<br />- Total Electric Flux = 4,640 volts-meters<br />- Total Area = Area of smaller square + Area of larger square = \( s^2 + 9s^2 = 10s^2 \)<br /><br />We can rearrange the formula to solve for the total electric flux:<br /><br />\[ 4,640 = 29.00 \times 10s^2 \]<br /><br />Solving for \( s^2 \):<br /><br />\[ s^2 = \frac{4,640}{29.00 \times 10} \]<br />\[ s^2 = \frac{4,640}{290} \]<br />\[ s^2 = 16 \]<br />\[ s = \sqrt{16} \]<br />\[ s = 4 \, \text{meters} \]<br /><br />Now, we need to find the electric flux through the larger square, which has an area of \( 9s^2 \):<br /><br />\[ \text{Electric Flux through Larger Square} = \text{Electric Field Strength} \times \text{Area of Larger Square} \]<br />\[ \text{Electric Flux through Larger Square} = 29.00 \times 9s^2 \]<br />\[ \text{Electric Flux through Larger Square} = 29.00 \times 9 \times 16 \]<br />\[ \text{Electric Flux through Larger Square} = 29.00 \times 144 \]<br />\[ \text{Electric Flux through Larger Square} = 4,176 \, \text{volts-meters} \]<br /><br />Therefore, the electric flux through the larger square is 4,176 volts-meters.