Motors draw lagging vars; capacitors supply leading ones. Put them on the same busbar and the vars cancel locally instead of being shipped down the utility's cable — the real power is untouched, and the hypotenuse of the triangle pulls in.

$Q_{c} = P\left(\tan\varphi_{1} - \tan\varphi_{2}\right)$, read aloud Q-c equals P times tan phi-one minus tan phi-two. $Q_{c}$ is the capacitor rating in kvar; $P$ is the load's real power in kilowatts, and it does not change; the subscripts are a before and an after — 1 is where the power factor is today, 2 is where you want it — so $\varphi_{1} = \arccos(\text{PF}_{1})$ and $\varphi_{2} = \arccos(\text{PF}_{2})$, both angles in degrees. The tangent is what converts a power factor into vars-per-watt; the cosine cannot do that job, and subtracting the two power factors themselves is the marquee mistake here.

Then size the component. $C = \dfrac{Q_{c}}{2\pi f V^{2}}$ — C equals Q-c over two pi f V squared — where $C$ is the capacitance in farads (a panel would quote it in microfarads), $f$ is the supply frequency in hertz and $V$ the RMS voltage across the capacitor in volts. Note that $V$ is squared: the same capacitor gives four times the vars at twice the voltage, which is why a unit is stamped with a voltage as firmly as with a kvar.

Two cautions worth carrying out of here. Correcting all the way to unity is rarely the target — overshoot on a light load leaves you LEADING, which the utility likes no better. And capacitors and harmonics are a poor pairing; on a plant full of drives, correction gets detuning reactors or it gets replaced.