# Physics explanation of charging time by increasing dielectric constant

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?

• I will answer your question when I am at school. I will also try to talk to my physics teacher. – domenix Feb 2 '16 at 4:39
• Try physics stack exchange. – Andy aka Feb 2 '16 at 8:41

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.

• Well, thanks for the answer. This is not a homework. I am not at school anymore. However, I think you misunderstood my question. I know that formula and how increase dielectric constant also increase time constant. But what I am stuck is how to explain it in physics. With dialectic material between two plates of capacitors, the electric field is reduced. But how that affect the time constant? (Not using time constant) – anhnha Feb 2 '16 at 4:19
• @user3126592 see my update. Is that what you were thinking? Otherwise it may have to start moving into the realm of differential equations. Or I could be completely off the mark as to what you are trying to find out. – Tom Carpenter Feb 2 '16 at 4:27
• You used the formula V = Q/C with two different capacitance C1, C2 and C1>C2. For example to reach a given voltage both capacitors have the same initial current but the larger one will need more charge and so it takes more time. If I understand you correctly that is your argument. But I think that is not complete. How would you exclude the case where the larger capacitor will have charging rate much much faster? So finally it will get the target voltage before the smaller one. I know in reality it is wrong but how can we explain this? – anhnha Feb 2 '16 at 4:51
• @user3126592 For that to be the case, the resistance in the R-C circuit would have to be lower for the larger capacitor. If that was the case then yes it would charge faster, but equally the time constant would be lower as $\tau=RC$. The RC constant can be derived as correct by analysis using differential equations. I can add that to the answer in the morning, but it's just gone midnight here, so need some rest. – Tom Carpenter Feb 2 '16 at 5:05
• Thank you. I can do it myself but I was stuck at understanding it in physics. – anhnha Feb 2 '16 at 19:12

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).