Electromagnetic Waves

JEE Main level, single correct, medium

6 of 165

Question 6 of 165

A displacement current of 4.04.0 A can be set up in the space between two parallel plates of 66 μ\muF capacitor. The rate of change of potential difference across the plates of the capacitor is nearly α×106\alpha \times 10^6 V/s. The value of α\alpha is __________.

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Answer: (B) 0.670.67

The displacement current IdI_d between the plates of a capacitor is given by the formula:

Id=CdVdtI_d = C \dfrac{dV}{dt}

Substituting the given values of displacement current and capacitance:

4.0=6×10−6×dVdt4.0 = 6 \times 10^{-6} \times \dfrac{dV}{dt}

dVdt=4.06×10−6\dfrac{dV}{dt} = \dfrac{4.0}{6 \times 10^{-6}}

dVdt=23×106≈0.67×106\dfrac{dV}{dt} = \dfrac{2}{3} \times 10^6 \approx 0.67 \times 10^6 V/s

Comparing this with the given rate of change of potential difference α×106\alpha \times 10^6 V/s, we get α=0.67\alpha = 0.67.

Answer: 0.670.67

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