A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance 2.5 μ F. The dielectric constant of the medium between the capacitor plates is 1 . It produces an instantaneous displacement current of 0.25 mA in the intervening space between the capacitor plates, the magnitude of the rate of change of the potential difference will be _________ Vs⁻¹.

Solution

Physics sir · 4 steps

Displacement current equals the charging current, so I_d = C dV/dt gives the answer in one line.

1. Displacement current. Between the plates, I_d = ε₀ dΦ_E/dt. For a parallel plate capacitor Φ_E = EA = VA/d, so I_d = (ε₀ A)/ddV/dt.
2. Spot the capacitance. With dielectric constant 1, the factor (ε₀ A)/d is the capacitance C itself.

   I_d = C dV/dt ⇒ dV/dt = I_d/C

3. Put in numbers. I_d = 0.25 mA = 2.5 × 10⁻⁴ A and C = 2.5 μF = 2.5 × 10⁻⁶ F. So dV/dt = (2.5 × 10⁻⁴)/(2.5 × 10⁻⁶) = 100 V s⁻¹.
4. So. The potential difference changes at 100 V s⁻¹.

The common slip: 0.1. Dividing 0.25 by 2.5 without converting mA and μF to SI gives 0.1, not 100.

The idea to keep. Inside a capacitor, the displacement current equals the conduction current in the wires: I_d = C dV/dt. Convert milli and micro to SI before you divide.

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