Displacement Current

The displacement current that Maxwell added is not carried by moving charges but by the rate of change of electric flux through a surface, written as i_d = \varepsilon_0 \frac{d\Phi_E}{dt}. In the insulating gap of a charging capacitor this quantity exactly matches the conduction current in the connecting wires, which is why the magnetic field stays continuous. Substituting \varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/\text{N·m}^2 and d\Phi_E/dt = 6.0 \times 10^{9}\ \text{V·m/s} gives i_d = 8.85 \times 10^{-12} \times 6.0 \times 10^{9} = 5.31 \times 10^{-2}\ \text{A}, which is about 53 mA. The value 5.3 mA is ten times too small and would follow from a flux rate of 6 \times 10^{8}. The value 0.53 A overstates the result by a factor of ten. The 530 mA option corresponds to an even larger arithmetic slip in the exponent. This calculation follows directly from the Ampere-Maxwell formulation in the NCERT Electromagnetic Waves chapter, and it shows that even a purely field-based quantity can be expressed cleanly in amperes. The same displacement current would also equal the conduction current charging the plates, so one could verify it independently from the rate at which charge accumulates if that data were given. As a check, multiplying a permittivity (\sim10^{-11}) by a flux rate (\sim10^{9}) should land near 10^{-2}\ \text{A}, consistent with the 53 mA answer.