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| ### Statement |
| ### Statement |
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| $11.2.20.$ [Insert the problem statement] |
| $11.2.20.$ |
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| The circuit of an electric circuit includes a resistance R and an uncharged |
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| capacitor of capacitance C. |
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| a. Prove that the charge on the capacitor during the appearance and then |
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| disappearance of the magnetic flux through the circuit does not exceed the |
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| valueΦT |
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| CR2, where T is the lifetime of this magnetic flux, Φ is its maximum |
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| value. |
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| b. To determine the direct current I flowing in the circuit during time T, if |
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| there is an alternating current caused by electromagnetic induction in the |
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| same time interval, measure the capacitance potential V after all the cur- |
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| rents disappear, and then estimate the direct current by the formula I =CV |
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| T |
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| Determine the maximum error of such an estimate in the case when the AC |
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| amplitude is k times greater than I. |
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| ### Solution |
| ### Solution |
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| 1. Fields in the capacitor at rest |
| firs analyse the Field in the capacitor at rest |
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| · Plate area: A. |
| Plate area: A. |
| · Separation: d. |
| Separation: d. |
| · Capacitance: C = \dfrac{\varepsilon_0 A}{d}. |
| Capacitance: $C = \dfrac{\varepsilon_0 A}{d}$. |
| · Charge: Q = CV. |
| Charge:$ Q = CV$. |
| · Surface density: \sigma = \dfrac{Q}{A} = \dfrac{CV}{A}. |
| Surface density: $\sigma = \dfrac{Q}{A} = \dfrac{CV}{A}$. |
| · Electric field between plates (from the positive to the negative): |
| · Electric field between plates (from the positive to the negative): |
| \mathbf{E}_0 = -\frac{V}{d}\,\hat{\mathbf{z}} \quad \text{(uniform)}. |
| $ \mathbf{E}_0 = -\frac{V}{d}\,\hat{\mathbf{z}} \quad \text{(uniform)}$. |
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| At rest, there is no magnetic field. |
| At rest, there is no magnetic field. |
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| Now ,at a given instant, the capacitor moves with velocity v (variable due to acceleration). We assume v \ll c (non‑relativistic). |
| Now ,at a given instant, the capacitor moves with velocity v (variable due to acceleration). We assume$ v \ll c$ (non‑relativistic). |
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| The moving charged plates constitute surface currents: |
| The moving charged plates constitute surface currents: |
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| · Upper plate (charge +Q) at z = d/2: \mathbf{K}_+ = \sigma v\,\hat{\mathbf{x}}. |
| 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}}. |
| · Lower plate $(charge -Q) at z = -d/2: \mathbf{K}_- = -\sigma v\,\hat{\mathbf{x}}$. |
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| 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: |
| 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: |
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| \mathbf{B} = \mu_0 \sigma v\,\hat{\mathbf{y}}. |
| $\mathbf{B} = \mu_0 \sigma v\,\hat{\mathbf{y}}$. |
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| Outside the plates, \mathbf{B} = 0. |
| Outside the plates, |
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| $\mathbf{B} = 0$. |
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| We calculate the electromagnetic momentum |
| We calculate the electromagnetic momentum |
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| 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): |
| The momentum density of the electromagnetic field is |
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| \mathbf{P}_{\text{em}} = \varepsilon_0 \int \mathbf{E} \times \mathbf{B} \, dV. |
| $\varepsilon_0(\mathbf{E} \times \mathbf{B})$. |
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| With \mathbf{E} = -\frac{V}{d}\hat{\mathbf{z}}, \mathbf{B} = \mu_0 \sigma v \,\hat{\mathbf{y}}: |
| Integrating over the volume between the plates $ (V = A d)$: |
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| \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}}. |
| $\mathbf{P}_{\text{em}} = \varepsilon_0 \int \mathbf{E} \times \mathbf{B} \, dV$. |
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| Substituting \sigma = CV/A and C = \varepsilon_0 A/d: |
| With$ \mathbf{E} = -\frac{V}{d}\hat{\mathbf{z}}, \mathbf{B} = \mu_0 \sigma v \,\hat{\mathbf{y}}:$ |
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| \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}}. |
| $\mathbf{E} \times \mathbf{B} = \left(-\frac{V}{d}\hat{\mathbf{z}}\right) \times (\mu_0 \sigma v \,\hat{\mathbf{y}}) = -\frac{V}{d}\mu_0 \sigma v \,(\hat{\mathbf{z}} \times \hat{\mathbf{y}}) = \frac{V}{d}\mu_0 \sigma v \,\hat{\mathbf{x}}$. |
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| \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}}. |
| Substituting |
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| Now the electromagnetic mass |
| $\sigma = CV/A $ and $ |
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| C = \varepsilon_0 A/d$: |
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| $\mathbf{E} \times \mathbf{B} = \frac{V}{d}\mu_0 \frac{CV}{A} v \,\hat{\mathbf{x}} = \mu_0 \frac{C V^2}{A d} v \,\hat{\mathbf{x}}$. |
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| $\mathbf{P}_{\text{em}} = \varepsilon_0 \left( \mu_0 \frac{C V^2}{A d} v \right) (A d) \,\hat{\mathbf{x}} = \varepsilon_0 \mu_0 C V^2 v \,\hat{\mathbf{x}} = \frac{C V^2}{c^2}\,v\,\hat{\mathbf{x}}$. |
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| Now the calculate ele2ctromagnetic mass |
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| The total momentum of the field is proportional to the instantaneous velocity. The electromagnetic mass is defined as the proportionality constant: |
| The total momentum of the field is proportional to the instantaneous velocity. The electromagnetic mass is defined as the proportionality constant: |
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| m_{\text{em}} = \frac{P_{\text{em}}}{v} = \frac{C V^2}{c^2}. |
| $m_{\text{em}} = \frac{P_{\text{em}}}{v} = \frac{C V^2}{c^2}$. |
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| 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. |
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| #### Answer |
| #### Answer |
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| [Insert a concise answer or boxed result] |
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| $\boxed{m_{\text{em}} = \frac{C V^2}{c^2}}$. |
