Правка разделов «Statement», «Solution», «Answer»

Alexphysics правка от 16 июн. 2026 г., 01:41

правка #18890 предыдущая #18888 ← раньше

en/11.2.20.md +1 −4

			### Statement

			$11.2.20.$

			The circuit of an electric circuit includes a resistance R and an uncharged
			capacitor of capacitance C.
			a. Prove that the charge on the capacitor during the appearance and then
			disappearance of the magnetic flux through the circuit does not exceed the
			valueΦT
			CR2, where T is the lifetime of this magnetic flux, Φ is its maximum
			value.
			b. To determine the direct current I flowing in the circuit during time T, if
			there is an alternating current caused by electromagnetic induction in the
		@@ -14,7 +14,6 @@Statement
			same time interval, measure the capacitance potential V after all the cur-
			rents disappear, and then estimate the direct current by the formula I =CV
			.
		−	T
			Determine the maximum error of such an estimate in the case when the AC
			amplitude is k times greater than I.

			### Solution

			firs analyse the Field 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$:

			$\mathbf{E} \times \mathbf{B} = \frac{V}{d}\mu_0 \frac{CV}{A} v \,\hat{\mathbf{x}} = \mu_0 \frac{C V^2}{A d} v \,\hat{\mathbf{x}}$.
		@@ -72,14 +71,12 @@Solution

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

		−	Now the calculate ele2ctromagnetic mass
		+	Now the calculate electromagnetic mass

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

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

		−
			#### Answer
		−

			$\boxed{m_{\text{em}} = \frac{C V^2}{c^2}}$.