Problem Restatement:

The problem involves calculating the electric flux through the larger of two squares given the total electric flux through the surface formed by both squares. Here's the step-by-step breakdown:

Given:

1. Electric flux ($\Phi_E$) is defined as: $\Phi_E = E \cdot A$ where:

   • $E$: Electric field strength = $29.00 \, \text{V/m}$
   • $A$: Area of the surface in square meters.

2. The surface consists of:

   • A smaller square with side length $x$.
   • A larger square with side length $3x$ (three times the side length of the smaller square).

3. Total electric flux through the surface: $\Phi_{\text{total}} = 4640 \, \text{V·m}.$

To Find:

The electric flux ($\Phi_{\text{large}}$) through the larger square.

Solution:

1. Calculate Areas:

   • Smaller square area: $A_{\text{small}} = x^2$
   • Larger square area: $A_{\text{large}} = (3x)^2 = 9x^2$

2. Total Electric Flux: The total flux is the sum of the flux through the smaller and larger squares: $\Phi_{\text{total}} = \Phi_{\text{small}} + \Phi_{\text{large}}$ Substituting for flux: $\Phi_{\text{total}} = E \cdot A_{\text{small}} + E \cdot A_{\text{large}}$ Factor $E$ out: $\Phi_{\text{total}} = E \cdot (A_{\text{small}} + A_{\text{large}})$

3. Substitute Areas: Using the areas calculated: $\Phi_{\text{total}} = E \cdot (x^2 + 9x^2)$ Simplify: $\Phi_{\text{total}} = E \cdot 10x^2$

4. Solve for $x^2$: Substitute the given values ($\Phi_{\text{total}} = 4640 \, \text{V·m}$ and $E = 29.00 \, \text{V/m}$): $4640 = 29 \cdot 10x^2$ Simplify: $x^2 = \frac{4640}{290} = 16 \, \text{m}^2.$

5. Flux Through Larger Square: The flux through the larger square is: $\Phi_{\text{large}} = E \cdot A_{\text{large}}$ Substitute $A_{\text{large}} = 9x^2$: $\Phi_{\text{large}} = E \cdot 9x^2$ Substitute $x^2 = 16$ and $E = 29$: $\Phi_{\text{large}} = 29 \cdot 9 \cdot 16$ Calculate: $\Phi_{\text{large}} = 4176 \, \text{V·m}.$

Final Answer:

The electric flux through the larger square is: $\boxed{4176 \, \text{V·m}}$

Would you like a deeper explanation of any step?

Related Questions:

1. How is the concept of electric flux related to Gauss's law?
2. How does the area factor into electric flux if the surface is tilted relative to the field?
3. What would the flux be if the field strength doubled?
4. How would the flux calculation change if the squares were rectangles instead?
5. Can electric flux be negative, and in what situations?

Tip:

Always remember that electric flux depends on the orientation of the surface relative to the electric field—perpendicular orientation maximizes flux.