Two parallel plates of area 2.70 cm by 2.70 cm are separated by 1.50 mm and carry charges ±0.708 nC. What is the electric field strength between the plates?

• A. 1×10^5 N/C
• B. 2×10^5 N/C
• C. 3×10^5 N/C
• D. 4×10^5 N/C

Answer: (B)

The field between large, oppositely charged parallel plates is determined by the surface charge density on each plate: E = σ/ε0, where σ = Q/A. This comes from the fact that each plate creates a uniform field, and between the plates these fields add.

Compute the area: A = (2.70 cm)^2 = (0.0270 m)^2 = 7.29×10^-4 m^2.

Surface charge density: σ = Q/A = (0.708×10^-9 C) / (7.29×10^-4 m^2) ≈ 9.71×10^-7 C/m^2.

Electric field between the plates: E = σ/ε0 ≈ (9.71×10^-7 C/m^2) / (8.854×10^-12 C^2/(N·m^2)) ≈ 1.10×10^5 N/C.

So the field is about 1.1×10^5 N/C, which is essentially 1×10^5 N/C in the given choices. Note that the separation (1.50 mm) does not affect this ideal field magnitude for large, parallel plates; it would affect the potential difference via V = E d (about 0.165 kV in this case).