capacitor ground or not?
What you have is technically correct. The voltage on the positive terminals of polarized caps must always be greater than the negative terminal. What voltage the negative terminal is at is not significant.

What it functionally does here may not be what you intend.

As shown, the LED will light when you apply power then dim out shortly after as the capacitor charges up. Subsequently it may never light again, or not for the very long time it takes for the capacitor to leak it's charge.

Further, if your LED is actually the car lights, the 1K resistor will not provide enough current to light them.

edit: here is a rough diagram of what you get once the capacitor charges up.  the voltage on both sides of the LED will be the close to the same, so no light







simulate this circuit &ndash; Schematic created using CircuitLab



For something like you are describing you would need something like this..

(Ignore the two switches.. they are just for testing in the schematic editor)







simulate this circuit

When you switch on the circuit C1 charges up through R1. When the voltage reaches the threshold voltage of the N_MOSFET the latter turns on which switches on the P-MOSFET turning on your lights. Delay time is set by the R1, C1 combination and also depends on the gate threshold of the N-MOSFET. You want the latter to be closer to 6V than 1V.

Making R1 a 1 Meg pot will allow you to adjust the delay.

When you open the switch, the capacitor will discharge, initially through the light till M2 turns off, then though the diode D1 and the smaller resistor R3.

R2 simply biases M1 off.

If you want to build this, you need to use MOSFETS that are designed to handle automobile transient voltages.

Here is  simpler version that only uses a P-MOSFET. It works sort of the same way as the previous one but the charge circuit is reversed. The switching edge with a single transistor is much slower though and may cause some flickering in the lights as they turn on.







simulate this circuit