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I am having a problem understanding in physics of the relationship between charging time and dielectric constant of a parallel plate capacitor. I'd like about physics aspect not based on time constant. If dielectric constant increases then capacitance also increases and so the time constant RC. However, how would you explain it in physics?
With dialectic material between two plates of capacitors, the electric field is reduced. But how that affect the time constant?
Capacitance of a parallel place capacitor is related to the dielectric constant, area, and distance between plates by a very simple equation:
$$C = \frac{\epsilon_r \epsilon_0 A}{d}$$
Where \$C\$ is capacitance, \$A\$ is plate area, and \$d\$ is the separation between plates. \$\epsilon_0\$ is the permittivity of free space and \$\epsilon_r\$ is the relative permittivity.
The voltage across a capacitor at any given instant is given by:
$$V = \frac{Q}{C}$$
Where \$V\$ is voltage, \$Q\$ is the charge on the plates, and \$C\$ is the capacitance.
So clearly as capacitance increases, the amount of charge required for a given voltage across the plates will also increase proportionally.
If you require more charge, it will take longer to reach a given voltage if you have the same initial current - which in an R-C circuit is governed by the resistor. In essence at time \$t=0\$, if there is no charge on the capacitor, all of the voltage is across the resistor, so this sets the maximum current. As the capacitor charges, the voltage across the resistor will decrease (it increases over the capacitance), so current decreases (Ohms law).
Given that current is the rate of change of charge, then with the same limitation on current, it will take longer to charge the larger capacitor to a given voltage.
Simply said: the effect of an electric field on a dielectric medium is a displacement of the electron clouds with respect to the atomic nuclei. Said effect is called polarization. This displacement requires additional coulombic force (and energy), therefore more charge is needed on the plates of the capacitor, in effect increasing it's capacitance. And increasing capacitance also increases charging time as long as you keep current fixed (i=C*du/dt).
