It is absolutely easy and comfortable to use the power MOSFET as a DC switch as in your case, it must be much easier than using a BJT with the better characteristics.
You need to calculate nothing, just care of the MOSFET specs itself. In most case, you need only a resistor of about 1M to 10M ohm between G and S (you may omit this resistor), a resistor of about 100 ohm to 1M ohm (meanings that you have a very wide range of free choice of resistors but always remember that these 2 resistors also make a voltage divider and your resulted Vgs may be lower than needed level).
About the MOSFET specs, you should know about the maximum Vds, Vgs, Id, and maybe Rds ON. Of course, for n-channel type, you have to ensure that voltage at G must be more positive in respect to S to make MOSFET works.
For most of MOSFET, Vgs = 5... 10V is good enough to make MOSFET open saturately. For logic-level MOSFET, it will work at logic-level of 5V, this type of MOSFET will have the letter L as the suffix of symbol, maybe your CEB6030L is that type, please check with its datasheet.
With your simple use, you can pick out the appropriate circuit from MOSFET datasheet to use.
Goodluck
nguyennam