BITSAT 2024 Shift 1 Paper

Given the radius of the capacitor plates $R=0.1\ \mathrm{m}$, we can calculate the area of the plates as:$A=\pi R^2=3.14\times(0.1)^2\ \mathrm{m}^2$We are also given that the rate of change of the electric field between the capacitor plates is:$\frac{dE}{dt}=6\times10^{13}\;\frac{\mathrm{V}}{\mathrm{m}\cdot\mathrm{s}}$The displacement current $I_d$ between the plates is calculated using the formula:$I_d=\varepsilon_0\;\frac{d\phi}{dt}$where $\frac{d\phi}{dt}=A\;\frac{dE}{dt}$. Hence, the formula becomes:$I_d=\varepsilon_0 A\left(\frac{dE}{dt}\right)$Substituting the values, where $\varepsilon_0=8.85\times10^{-12}\ \mathrm{F}/\mathrm{m}$, we have:$I_d=8.85\times10^{-12}\times3.14\times(0.1)^2\times6\times10^{13}$Calculating the above expression yields:$I_d=16.67\ \mathrm{A}$Therefore, the value of the displacement current is 16.67 A .