New solution

Alexphysics edited Jun 16, 2026, 1:22 AM

View solution History

revision #18886 newer →

en/11.2.20.md +59 −0

		@@ -0,0 +1,59 @@
		+	### Statement
		+
		+	$11.2.20.$ [Insert the problem statement]
		+
		+	### 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.
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,59 @@
			### Statement

			$11.2.20.$ [Insert the problem statement]

			### 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.

			#### Answer

			[Insert a concise answer or boxed result]