# Doubling the voltage to a capacitor doubles the stored charge but does not make it charge slower. Why not?

This is a slightly theoretical question.

If you charge a capacitor from say a 3V battery you get Q charge in T seconds. If you double the voltage to 6V you get 2Q charge also in T seconds.

My question is, why doesn't T change? If you think of a capacitor like a room and charge like people, if you want to fit in twice as many people, surely you need more time?

I am thinking of the formula Q1 = Q0e^-(t/RC). Is there something I am missing?

• You have a higher voltage 'pushing' the charge in. Jan 30, 2018 at 7:22
• You opened two doors this time Jan 30, 2018 at 8:22

Current is rate of charge (1 ampere is 1 coulomb/sec). At time zero, the capacitor voltage is zero, so the voltage across your resistance is three volts. The instantaneous current is 3V/R. When you apply 6V (assuming the resistance is equal and ignoring @Whit3rd 's excellent point that adding a second battery will likely increase the resistance), the instantaneous current at time zero would be 6V/R, twice that of the first case. This means that in your analogy, the people would be walking into the room in the first case, and running in the second. In the second case, when the capacitor reaches 3 volts, the voltage across your resistor is (6-3) volts, and the current would be equal to the starting current with the 3 volt supply. At this point the people in the second case would have slowed down and be walking in at the same speed as the people in the first case had started.

charge a capacitor from say a 3V battery you get Q charge in T seconds. If you double the voltage to 6V you get 2Q charge also in T seconds

Actually, probably you don't. Depends on 'R'.

$$Q = C \times V_{capacitor}$$ $$V_{capacitor} = V_{battery} \times (1- e ^ {-T \over{R \times C}})$$

If your '3V' source is a battery, with some internal series resistance, and puts Q charge on an uncharged capacitor in a short time T, then a '6V' source might be two batteries in series, with twice the internal resistance, and will put 2Q charge on a capacitor in time 2T, through resistance 2R.

In very long time scales (times much greater than the product of resistance and capacitance), the capacitor and battery circuit can be deemed to have negligible resistance (because the current times resistance is very small when the capacitor is near fully charged), but that is an approximation that does NOT allow a prediction of time versus voltage, only allows the prediction of the at-equilibrium charge.

In a time model, you NEED to be explicit that there is a resistor, but if none is 'designed' in, the resistance comes from stray effects. Batteries' internal resistance when placed in series, for example.

The equation for capacitor is:

Q = C V

Then you can apply derivative over t, that is d/dt, for both sides of the equation to get:

dQ/dt = C dV/dt

I = C dV/dt

So over time T you may get as many charge Q as you want: dQ/dT. It just depends on the current you are using to charge your capacitor.

I encourage you to read the articles here:

You can see how a capacitor is charged and discharged by different currents on these pictures (look into articles):  Hope it will clarify you the usage of the capacitor and the relation between its charge Q and the change of it voltage V over time T (dV/dT).