In this example
After the initial charging of the cap to 3V, current gets blocked, but over time does it consume any energy from the batteries? Is this safe to make?
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Leakage current will drain the battery, most likely not that significantly compared to the internal self-discharge of the battery.
An aluminum electrolytic might leak 100nA long term, which is not much compared to the self-discharge of even a button cell. The guaranteed maximum of a typical e-cap of this size is 0.002CV or 400nA (whichever is greater) after 3 minutes. Most parts will beat that significantly. Some SMD parts are not nearly as good.
Your second question was whether this safe to make. Generally, yes, however there are almost always exceptions in engineering. If your 3V battery has a large current capacity (perhaps an unprotected 18650 Li cell) and your capacitor is something like a 6.3V tantalum capacitor there is a significant risk of an 'ignition' event upon connecting the capacitor to the battery (picture flames shooting out, a bright light and some noxious fumes). The risk can be considerably reduced by adding some series resistance of some tens of ohms.
In steady state (after a long time) an ideal capacitor does not draw significant current from a battery. A real capacitor will draw some small leakage current. The amount of leakage current will depend on the type of the capacitor, electrolytics will have higher leakage than films and ceramics.
An ideal capacitor would be open circuit to DC, so no current would flow, and no energy would be consumed after the capacitor is fully charged.
However, real capacitors do have some small leakage current, so, in Real Life, energy would be consumed from the battery very slowly after the initial charging.
You should check something called "insulation resistance"
I quote from Murata:
The insulation resistance of a monolithic ceramic capacitor represents the ratio between the applied voltage and the leakage current after a set time (ex. 60 seconds) while applying DC voltage without ripple between the capacitor terminals. While the theoretical value of a capacitor's insulation resistance is infinite, since there is less current flow between insulated electrodes of an actual capacitor, the actual resistance value is finite. This resistance value is called "insulation resistance" and denoted with units such as Meg Ohms [MΩ] and Ohm Farads [ΩF].
I checked a datasheet I had (part number: GRM32ER71H106KA12) for an example to approximate how much leakage gets to pass. Check the image below:
To fully understand the behavior of the capacitor at steady state (as in directly connecting a capacitor to a battery) I highly recommend reading the following article: http://www.murata.com/support/faqs/products/capacitor/mlcc/char/0003
If the polarity of the battery is reversed in this scienario, then even an ideal capacitor will consume current to change it's polarity in tune with the battery. But in this case only a real capacitor will be able to consume energy due to springing effect i.e leakage of charge from the edges of the capacitor. However it'll depend on the type of capacitor and the material used in making the capacitor.