The displacement current of 4.425 μA is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 10⁶ Vs⁻¹. The area of each plate of the capacitor is 40 cm². The distance between each plate of the capacitor is x × 10⁻³ m. The value of x is __________.
(Permittivity of free space, E₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²).

Solution

Physics sir · 4 steps

Displacement current is just C times dV/dt. Put in C = ε₀A/d and solve for the gap d.

1. Displacement current. Between the plates, the displacement current equals the conduction current in the wires. So I_d = CdV/dt, and for a parallel plate capacitor C = (ε₀ A)/d.

   I_d = (ε₀ A)/d dV/dt ⇒ d = (ε₀ A)/I_d dV/dt

2. Convert units. A = 40 cm² = 40 × 10⁻⁴ m² = 4 × 10⁻³ m² and I_d = 4.425 × 10⁻⁶ A.
3. Put in numbers. The numerator is 8.85 × 10⁻¹² × 4 × 10⁻³ × 10⁶ = 3.54 × 10⁻⁸. Dividing, d = (3.54 × 10⁻⁸)/(4.425 × 10⁻⁶) = 8 × 10⁻³ m.
4. So. The plate separation is 8 × 10⁻³ m, so x = 8.

The common slip: 800. Converting 40 cm² as 40 × 10⁻² m² gives d = 0.8 m, which is 800, not 8. One square centimetre is 10⁻⁴ m².

The idea to keep. Displacement current between capacitor plates is ε₀ dΦ_E/dt = CdV/dt. Any question on it is really a capacitance question in disguise.

Answer from the NTA final answer key. Solution by PrepWiser.