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Electromagnetic Waves question

2025 · 23 Jan · Shift 2 · Q71
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Electromagnetic Waves question

2025 · 23 Jan · Shift 2 · Q71

JEE MainPhysicsElectromagnetic WavesNumerical+4 / −1
A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance 2.5μ F2.5 \mu \mathrm{~F}2.5μ 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\underline{\hspace{2cm}}\mathrm{Vs}^{-1}​Vs−1.
Numerical answer
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Correct answer: 100

  1. For a capacitor, the displacement current is equal to the capacitor current:

Id=C dVdtI_d = C\,\frac{dV}{dt}Id​=CdtdV​

  1. Given:

C=2.5 μF=2.5×10−6 FC = 2.5\,\mu F = 2.5 \times 10^{-6}\,FC=2.5μF=2.5×10−6F

Id=0.25 mA=0.25×10−3 AI_d = 0.25\,mA = 0.25 \times 10^{-3}\,AId​=0.25mA=0.25×10−3A

  1. Therefore,

dVdt=IdC\frac{dV}{dt} = \frac{I_d}{C}dtdV​=CId​​

Substitute the values:

dVdt=0.25×10−32.5×10−6\frac{dV}{dt} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}}dtdV​=2.5×10−60.25×10−3​

  1. Simplify:

dVdt=0.252.5×103=0.1×103=100 V s−1\frac{dV}{dt} = \frac{0.25}{2.5} \times 10^{3} = 0.1 \times 10^{3} = 100\,V\,s^{-1}dtdV​=2.50.25​×103=0.1×103=100Vs−1

  1. Hence, the magnitude of the rate of change of potential difference is:

100 V s−1\boxed{100\,V\,s^{-1}}100Vs−1​

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