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When a capacitor is charging, the rate of change $dE/dt$ of the electric field between the plates is non-zero, and from the Maxwell-Ampère equation this causes a circulating magnetic field.

Now, since a magnetic field exists, why is the energy of a capacitor only stored in the electric field? Usually the formula for the energy stored goes as $ W = \pi d A \times \frac{1}{2}\epsilon_0 E^2$, where the first term is the volume and latter is the electric field energy density.

In Poynting's theorem, the electro-magnetic field energy density is $ \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0} B^2 $, i.e. there is also the magnetic field B present.

In a capacitor B is non-zero, so why do we not include it in the calculation of the energy stored?

In other words, why is the energy stored in a capacitor just $[ (volume) \times \frac{1}{2}\epsilon_0 E^2 ] $ and not $[ (volume) \times (\frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0} B^2) ] $ ?

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You are correct, that while charging a capacitor there will be a magnetic field present due to the change in the electric field. And of course $B$ contains energy as pointed out. However: As the capacitor charges, the magnetic field does not remain static. This results in electromagnetic waves which radiate energy away. The energy put into the magnetic field during charging is lost in the sense that it cannot be feed back to the circuit by the capacitor.

In the limit of a fully charged capacitor, there is no displacement current maintaining a magnetic field and all the energy is stored in the electric field.

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Radiation of EM energy is negligible in common circuits with low frequency currents. Magnetic energy in quasi-static cases like this transforms into electrical energy or dissipates as Joule heat in the wires. –  Ján Lalinský Feb 13 at 19:16

For a constant potential on the capacitor, there is no B-field and that is the case usually considered for this calculation. When charging a capacitor, the currents will generate a B-field and there is stored energy in that field (same as for an inductor). But once the charging stops, the B-field will "collapse" and cause currents to flow in the wires, dissipating that energy. Real capacitors will have some inductance and so will the wires feeding the capacitor and yes, you might need to include the effects if they are large enough (and they often do get included when analyzing circuits with real, not ideal, components).

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