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 .

There is 3phase power cable from 1 point to the other point
which is 100m away. Two point equipped with power meter and
recorded 1st point is 205kW while the 2nd is 202kW and each
phase average carry 400amps. To calculate the resistance of
the 3phase power cable, is the formula using will be P=I2R?