Question

21 mark for review 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 be $x$ meters.

Then side - length of larger square is $3x$ meters.

Step2: Calculate the areas of the two squares.

Area of smaller square $A_1=x^{2}$ square - meters, area of larger square $A_2=(3x)^{2}=9x^{2}$ square - meters. Total area $A = A_1+A_2=x^{2}+9x^{2}=10x^{2}$ square - meters.

Step3: Use the electric - flux formula $\varPhi = E\times A$.

We know that $\varPhi = 4640$ volts $\cdot$ meters and $E = 29$ volts per meter. Since $\varPhi=E\times A$, then $A=\frac{\varPhi}{E}$. Substituting the given values, we get $A=\frac{4640}{29}=160$ square - meters.

Step4: Find the value of $x$.

Since $A = 10x^{2}=160$, then $x^{2}=16$, and $x = 4$ meters.

Step5: Calculate the area of the larger square.

The area of the larger square $A_2=9x^{2}$. Substituting $x = 4$, we get $A_2=9\times16 = 144$ square - meters.

Step6: Calculate the electric flux through the larger square.

Using $\varPhi_2=E\times A_2$, with $E = 29$ volts per meter and $A_2 = 144$ square - meters, we have $\varPhi_2=29\times144 = 4176$ volts $\cdot$ meters.

Answer:

4176