Solution

1. Fields in the capacitor at rest

· Plate area: A.
· Separation: d.
· Capacitance: C = \dfrac{\varepsilon_0 A}{d}.
· Charge: Q = CV.
· Surface density: \sigma = \dfrac{Q}{A} = \dfrac{CV}{A}.
· Electric field between plates (from the positive to the negative):
\mathbf{E}_0 = -\frac{V}{d}\,\hat{\mathbf{z}} \quad \text{(uniform)}.

At rest, there is no magnetic field.

Now ,at a given instant, the capacitor moves with velocity v (variable due to acceleration). We assume v \ll c (non‑relativistic).

The moving charged plates constitute surface currents:

· Upper plate (charge +Q) at z = d/2: \mathbf{K}_+ = \sigma v\,\hat{\mathbf{x}}.
· Lower plate (charge -Q) at z = -d/2: \mathbf{K}_- = -\sigma v\,\hat{\mathbf{x}}.

These currents generate a magnetic field between the plates. Using Ampère's law for each plane and superposing, we obtain a uniform field in the region between the plates:

\mathbf{B} = \mu_0 \sigma v\,\hat{\mathbf{y}}.

Outside the plates, \mathbf{B} = 0.

We calculate the electromagnetic momentum

The momentum density of the electromagnetic field is \varepsilon_0(\mathbf{E} \times \mathbf{B}). Integrating over the volume between the plates (V = A d):

\mathbf{P}_{\text{em}} = \varepsilon_0 \int \mathbf{E} \times \mathbf{B} \, dV.

With \mathbf{E} = -\frac{V}{d}\hat{\mathbf{z}}, \mathbf{B} = \mu_0 \sigma v \,\hat{\mathbf{y}}:

\mathbf{E} \times \mathbf{B} = \left(-\frac{V}{d}\hat{\mathbf{z}}\right) \times (\mu_0 \sigma v \,\hat{\mathbf{y}}) = -\frac{V}{d}\mu_0 \sigma v \,(\hat{\mathbf{z}} \times \hat{\mathbf{y}}) = \frac{V}{d}\mu_0 \sigma v \,\hat{\mathbf{x}}.

Substituting \sigma = CV/A and C = \varepsilon_0 A/d:

\mathbf{E} \times \mathbf{B} = \frac{V}{d}\mu_0 \frac{CV}{A} v \,\hat{\mathbf{x}} = \mu_0 \frac{C V^2}{A d} v \,\hat{\mathbf{x}}.

\mathbf{P}_{\text{em}} = \varepsilon_0 \left(\mu_0 \frac{C V^2}{A d} v\right) (A d)\,\hat{\mathbf{x}} = \varepsilon_0 \mu_0 C V^2 v \,\hat{\mathbf{x}} = \frac{C V^2}{c^2}\,v\,\hat{\mathbf{x}}.

Now the electromagnetic mass

The total momentum of the field is proportional to the instantaneous velocity. The electromagnetic mass is defined as the proportionality constant:

m_{\text{em}} = \frac{P_{\text{em}}}{v} = \frac{C V^2}{c^2}.

Observation: Although the motion is accelerated, the expression for the fields and the momentum as a function of v is instantaneous. As long as v \ll c, m_{\text{em}} is constant. If radiation effects due to acceleration had to be considered, a reaction force term would appear, but the inertial mass of electromagnetic origin remains C V^2 / c^2.