A parallel-plate capacitor consists of two square plates of side 4.0 cm separated by 2.2 mm, with an electric field inside of 1.0×10^6 N/C. What is the charge on the positive plate?

• A. +14.2 nC
• B. +1.42 nC
• C. -14.2 nC
• D. +142 nC

Answer: (A)

The electric field between parallel plates is related to the surface charge density by E = σ/ε0, where σ = Q/A. So the charge on the plate is Q = ε0 E A.

Compute the area: each side is 4.0 cm = 0.04 m, so A = (0.04 m)^2 = 1.6×10^-3 m^2.

Using ε0 ≈ 8.854×10^-12 F/m and E = 1.0×10^6 N/C,

Q = (8.854×10^-12)(1.0×10^6)(1.6×10^-3) C ≈ 1.42×10^-8 C ≈ 14.2 nC.

The positive plate therefore carries about +14.2 nC. The given separation isn’t needed beyond assuming a uniform field; the result follows from Q = ε0 E A.