For an electric field passing through a flat surface perpendicular to it, the electric flux of the electric field through...
GMAT Problem-Solving and Data Analysis : (PS_DA) Questions
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?
1. TRANSLATE the problem information
- Given information:
- Two adjacent squares make up the flat surface
- Larger square side length = 3 × smaller square side length
- Electric field strength = 29.00 volts per meter
- Field passes uniformly and perpendicularly through surface
- Total electric flux = 4,640 volts·meters
- Electric flux = electric field strength × area
- What this tells us: We need to find individual areas to determine the flux through the larger square specifically.
2. INFER the strategic approach
- Key insight: If we let \(\mathrm{s}\) = side length of smaller square, then:
- Smaller square area = \(\mathrm{s^2}\)
- Larger square area = \(\mathrm{(3s)^2 = 9s^2}\)
- Total area = \(\mathrm{s^2 + 9s^2 = 10s^2}\)
- Strategy: Use the total flux equation to find \(\mathrm{s}\), then calculate the larger square's individual flux.
3. SIMPLIFY to find the smaller square's side length
- Set up equation using total flux:
\(\mathrm{4,640 = 29.00 \times 10s^2}\)
\(\mathrm{4,640 = 290s^2}\)
\(\mathrm{s^2 = 4,640 \div 290 = 16}\)
\(\mathrm{s = 4\text{ meters}}\)
4. Calculate the larger square's area and flux
- Area of larger square = \(\mathrm{9s^2 = 9(16) = 144\text{ square meters}}\)
- Electric flux through larger square = \(\mathrm{29.00 \times 144 = 4,176\text{ volts·meters}}\)
Answer: 4,176
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skills: Students misinterpret "3 times the side length" and incorrectly think the larger square has 3 times the area (instead of 9 times the area).
They might set up: Total area = \(\mathrm{s^2 + 3s^2 = 4s^2}\), leading to \(\mathrm{s^2 = 4,640 \div (29 \times 4) = 40}\). Then they calculate the larger square flux as \(\mathrm{29 \times (3 \times 40) = 3,480\text{ volts·meters}}\). This leads to confusion and guessing among the available answer choices.
Second Most Common Error:
Poor INFER reasoning: Students correctly find that the total area is \(\mathrm{10s^2}\) and solve for \(\mathrm{s^2 = 16}\), but then mistakenly calculate the flux through the larger square using only the scaling factor: \(\mathrm{4,640 \times (9/10) = 4,176}\). While this gives the right answer, it's conceptually incorrect reasoning that could fail in similar problems.
The Bottom Line:
This problem requires careful attention to how linear scaling affects area (squares the scaling factor) and systematic application of the flux formula at each step.