Question

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 fields 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?

Explanation:

Step1: Let side - length of smaller square

Let the side - length of the smaller square be $x$ meters. Then the side - length of the larger square is $3x$ meters.

Step2: Calculate areas of the squares

The area of the smaller square $A_1=x^{2}$ square meters, and the area of the larger square $A_2=(3x)^{2}=9x^{2}$ square meters. The total area of the surface $A = A_1+A_2=x^{2}+9x^{2}=10x^{2}$ square meters.

Step3: Use the electric - flux formula

The electric - flux formula is $\varPhi = E\times A$, where $\varPhi$ is the electric flux, $E$ is the electric - field strength, and $A$ is the area. Given $E = 29$ volts/meter and $\varPhi=4640$ volts - meters. We know that $\varPhi = E\times A$, so $4640=29\times10x^{2}$.

Step4: Solve for $x^{2}$

First, divide both sides of the equation $4640 = 29\times10x^{2}$ by $29\times10$:
\[x^{2}=\frac{4640}{29\times10}=\frac{4640}{290} = 16\]

Step5: Calculate the area of the larger square

The area of the larger square $A_2 = 9x^{2}$. Since $x^{2}=16$, then $A_2=9\times16 = 144$ square meters.

Step6: Calculate the electric flux through the larger square

Using the electric - flux formula $\varPhi_2=E\times A_2$. Substitute $E = 29$ volts/meter and $A_2 = 144$ square meters. Then $\varPhi_2=29\times144 = 4176$ volts - meters.

Answer:

4176