A parallel-plate capacitor has plates of area 25 cm^2 separated by 3.0 mm. The charge on the capacitor is 9.8 nC. A proton is released from rest next to the positive plate. How long does it take to reach the negative plate?

• A. 1.14×10^-8 s
• B. 1.00×10^-9 s
• C. 2.00×10^-8 s
• D. 5.00×10^-8 s

Answer: (A)

A proton released in a uniform electric field accelerates toward the negative plate with a constant acceleration a = qE/m. For parallel plates, the field between them is essentially uniform and is E = σ/ε0, where σ = Q/A.

Compute the surface charge density: σ = Q/A = (9.8e-9 C) / (25 cm^2). Convert area: 25 cm^2 = 2.5e-3 m^2, so σ ≈ 9.8e-9 / 2.5e-3 ≈ 3.92e-6 C/m^2. Then E ≈ σ/ε0 ≈ (3.92e-6) / (8.854e-12) ≈ 4.43e5 V/m.

Now the proton’s acceleration: a = qE/m, with q = 1.60e-19 C and m = 1.67e-27 kg. So a ≈ (1.60e-19 × 4.43e5) / 1.67e-27 ≈ 4.2e13 m/s^2.

The proton starts from rest and travels the plate separation d = 3.0 mm = 3e-3 m under constant acceleration, so t = sqrt(2d / a) ≈ sqrt(2 × 3e-3 / 4.2e13) ≈ sqrt(1.4e-16) ≈ 1.2e-8 s.

This gives about 1.1×10^-8 s, which matches the provided result.