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QJEEM25Ph74b4: JEE Main 2025 Physics - Class 12 JEEM25Ph74b4 Solution: A time varying potential difference is applied between the…

A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance 2.5μ F2.5 \mu \mathrm{~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\mathrm{Vs}^{-1} .

Step-by-Step Verified Solution

dVdt=IC\frac{dV}{dt} = \frac{I}{C}

Given that the displacement current is I=0.25×10−3 AI = 0.25 \times 10^{-3} \, \text{A} and the capacitance is C=2.5×10−6 F,C = 2.5 \times 10^{-6} \, \text{F}, the rate of change of the potential difference is calculated as follows:

dVdt=0.25×10−32.5×10−6=0.252.5×10−310−6=0.1×103=100 V/s.\frac{dV}{dt} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}} = \frac{0.25}{2.5} \times \frac{10^{-3}}{10^{-6}} = 0.1 \times 10^3 = 100 \, \text{V/s}.

Thus, the magnitude of the rate of change of the potential difference is 100 V/s100 \, \text{V/s} .

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