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NEET PHYSICSEasy

To produce an instantaneous displacement current of 2 mA2 \text{ mA} in the space between the parallel plates of a capacitor of capacitance 4 μF4 \text{ }\mu\text{F}, the rate of change of applied variable potential difference (dVdt)\left(\frac{dV}{dt}\right) must be:

A

800 V/s800 \text{ V/s}

B

500 V/s500 \text{ V/s}

C

200 V/s200 \text{ V/s}

D

400 V/s400 \text{ V/s}

Step-by-Step Solution

  1. Concept: Displacement current (IdI_d) arises due to a changing electric field (or electric flux) and is mathematically equivalent to the conduction current (IcI_c) flowing through the connecting wires of the capacitor.
  2. Formula: The relationship between displacement current, capacitance (CC), and the rate of change of voltage (dV/dtdV/dt) is derived from Q=CVQ = CV. Differentiating with respect to time gives I=dQdt=CdVdtI = \frac{dQ}{dt} = C \frac{dV}{dt}. Thus, Id=CdVdtI_d = C \frac{dV}{dt}.
  3. Given Data:
  • Displacement Current, Id=2 mA=2×10−3 AI_d = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}.
  • Capacitance, C=4 μF=4×10−6 FC = 4 \text{ }\mu\text{F} = 4 \times 10^{-6} \text{ F}.
  1. Calculation: dVdt=IdC\frac{dV}{dt} = \frac{I_d}{C} dVdt=2×10−34×10−6\frac{dV}{dt} = \frac{2 \times 10^{-3}}{4 \times 10^{-6}} dVdt=0.5×103=500 V/s\frac{dV}{dt} = 0.5 \times 10^3 = 500 \text{ V/s}
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