• $summary_{type}$: application
• $word_{count}$: 150

For a parallel plate capacitor being charged: $I_d$ = $epsilon_0$ * d$\frac{Phi_E}{dt}$ = $epsilon_0$ * A * dE/dt. Since E = $\frac{Q}{epsilon_0*A}$ between the plates: $I_d$ = $\frac{dQ}{dt}$ = $I_c$. The displacement current density $j_d$ = $epsilon_0$ * dE/dt (A/$m^2$). Between the plates, the magnetic field at radius r from the axis: B = $mu_0$$I_d$$\frac{r}{2*pi*R^2}$ for r < R (inside the plates), and B = $mu_0$$\frac{I_d}{2*pi*r}$ for r > R (same as a wire carrying $I_c$). Here R is the plate radius. If the capacitor voltage is V(t) = $V_0$sin(omegat): E = $\frac{V}{d}$, dE/dt = $V_0$omegacos$\frac{omega*t}{d}$, and $I_d$ = $epsilon_0$A$V_0$omegacos$\frac{omega*t}{d}$ = C$V_0$omegacos(omega*t) (since C = $epsilon_0$*A/d). JEE may ask for $I_d$ given dV/dt or dE/dt.