1.Existing P.F of Load (tanQ1)(0.92)
2.Required P.F (Improvement in P.F)-(tanQ2)(0.99)
3.Supply Voltage - for reference(Voltage variation if any)
4.Load in KW (exam- availabl with you i.e 500KW)

KVAR= KW(tanQ1-tanQ2)
KVAR= 500(tan0.92- tan0.99)
KVAR = 500(0.2890)
KVAR = 141.75 KVAR (@ 142KVAR)
if you want to calculate by amp need following data
efficiency of motor, required P.F, KW,maganatising current
of motor,formula find in Electrical Machine Engineering

250 Kvar Capacitor bank is required for 500Kw load
(i.e,625Kva).
This is because the capacitor rating for any load is equal
to 40% of Kva.
i.e, Capacitor required=40%x Kva Load

plz identify your systems existing power factor and
targeted power factor which u want to improve, then apply
this formula
KW X (Tan phi 1 - tan phi 2)

Reactive power (KVAR) is increase due to run inductive load.as a result power factor is lagging. so have to use capacitive power to reduce this inductive or reactive load. generally we have to use 63% capacitive load to remove the reactive load.i.e.

active power=500KW
so 60% of this will be=(60*500/100) = 300 KVAR

so we have to use 300KVAr capacitive capacitor bank to remove this reactive power .

If Three Runs of 1Cx300 Sqmm Cable between Transformer and VCB and the distance between 100mtrs. what will be total length of Cable and if cost 1500Rs per meter,what would be the total cost