Electronic – Capacitor across an ideal current source

I think the simulator is wrong. I would expect the capacitor to be charged to about 4.3 volts, allowing for an 0.7 Volt base to emitter drop in the transistor. (Although the transistor may fail due to excessive base current when the switch is first closed.)

In a perfect world, the capacitor would charge instantly. This is clear from your equation: the charge time is $$ t \approx 5RC $$ so if \$R = 0\$, then \$t = 0\$.

However, batteries are not perfect voltage sources. They have an effective resistance, which is on the order of 1 ohm, so the time to charge your capacitor without a resistor is approximately $$ t_{real} \approx 5C $$ This resistance depends on what type of battery, how dead the battery is, etc... so this is only a rough estimate.