Edits to “Statement”, “Solution”, “Answer”

Alexphysics edited Jun 17, 2026, 6:10 AM

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revision #18972 parent #18971 ← older newer →

en/14.3.11.md +14 −10

		@@ -1,13 +1,15 @@
			### Statement

		−	$14.3.11.$ ~~[Insert the problem statement]~~
		+	$14.3.11.$

		+	How many times will the potential difference and capacitance of a long cylin-drical capacitor change when it moves at a speed βc along the axis?
		+
			### Solution

			Reference frame and geometry

		−	· S': system at rest with respect to the capacitor.
		−	· S: laboratory system, in which the capacitor moves with velocity v along the z-axis (axis of the cylinder).
		+	S': system at rest with respect to the capacitor.
		+	S: laboratory system, in which the capacitor moves with velocity v along the z-axis (axis of the cylinder).

			The capacitor consists of two coaxial cylinders of radii a and b (b > a) and proper length L' (very large compared to the radii). In S' the electric field is purely radial and there is no magnetic field.

			For a boost along z, the components of the fields transform according to:

			$\begin{aligned}
			E_\parallel' &= E_\parallel, \quad B_\parallel' = B_\parallel,\\
			\mathbf{E}_\perp' &= \gamma (\mathbf{E}_\perp + \mathbf{v} \times \mathbf{B}),\\
			\mathbf{B}_\perp' &= \gamma \left(\mathbf{B}_\perp - \frac{\mathbf{v}}{c^2} \times \mathbf{E}\right).
			\end{aligned}$
		@@ -21,28 +23,28 @@Solution

			In our case, in S' (at rest):

		−	~~$\mathbf{E}'~~ is radial (perpendicular to z), therefore \mathbf{E}_\perp' = \mathbf{E}', E_\parallel' = 0$
		+	$\mathbf{E}'$
		+	is radial (perpendicular to z), therefore $\mathbf{E}_\perp' = \mathbf{E}', E_\parallel' = 0$
			$\mathbf{B}' = 0$

			In S
		−	using the inverse transformation (changing v \to -v)
		−	:
		+	using the inverse transformation (changing v \to -v) :

			$\mathbf{E}_\perp = \gamma (\mathbf{E}_\perp' - \mathbf{v} \times \mathbf{B}') = \gamma \mathbf{E}_\perp', \quad \mathbf{E}_\parallel = 0$

		−	$\mathbf{B}_\perp = \gamma \left(\mathbf{B}_\perp' + \frac{\mathbf{v}}{c^2} \times \mathbf{E}'\right) = \gamma \frac{\mathbf{v}}{c^2} \times \mathbf{E}'$.
		+	$\mathbf{B}_\perp = \gamma \left(\mathbf{B}_\perp' + \frac{\mathbf{v}}{c^2} \times \mathbf{E}'\right) = \gamma \frac{\mathbf{v}}{c^2} \times \mathbf{E}'$

			The radial electric field in S is:

			$\boxed{E_r = \gamma E_r'}$

			Potential difference between the cylinders

			The potential difference in S' is $
		−	V' = \int_a^b E_r' \, dr$.
		+	V' = \int_a^b E_r' \, dr$
			In S:

		−	$V = \int_a^b E_r \, dr = \int_a^b \gamma E_r' \, dr = \gamma V'$.
		+	$V = \int_a^b E_r \, dr = \int_a^b \gamma E_r' \, dr = \gamma V'$

			The transverse distances (a and b) do not change because the motion is parallel to the axis.

			$\boxed{V = \gamma V' = \frac{V'}{\sqrt{1-\beta^2}}}$.

			The potential difference increases by a factor of $ \gamma$

			Capacitance

			The total charge Q on the plates is a relativistic invariant (scalar). In S' the capacitance is $C' = Q/V'$
			In S:

			$C = \frac{Q}{V} = \frac{Q}{\gamma V'} = \frac{C'}{\gamma}$

			$\boxed{C = C' \sqrt{1-\beta^2}}$

			The capacitance decreases by a factor of \gamma (it is multiplied by
			$\sqrt{1-\beta^2})$
		@@ -64,4 +66,6 @@Solution

			#### Answer

		−	[Insert a concise answer or boxed result]
		+	$\boxed{C = C' \sqrt{1-\beta^2}}$
		+
		+	$\boxed{V = \gamma V' = \frac{V'}{\sqrt{1-\beta^2}}}$.

	@@ -1,13 +1,15 @@
	### Statement		### Statement

	$14.3.11.$ ~~[Insert the problem statement]~~		$14.3.11.$

			How many times will the potential difference and capacitance of a long cylin-drical capacitor change when it moves at a speed βc along the axis?

	### Solution		### Solution

	Reference frame and geometry		Reference frame and geometry

	· S': system at rest with respect to the capacitor.		S': system at rest with respect to the capacitor.
	· S: laboratory system, in which the capacitor moves with velocity v along the z-axis (axis of the cylinder).		S: laboratory system, in which the capacitor moves with velocity v along the z-axis (axis of the cylinder).

	The capacitor consists of two coaxial cylinders of radii a and b (b > a) and proper length L' (very large compared to the radii). In S' the electric field is purely radial and there is no magnetic field.		The capacitor consists of two coaxial cylinders of radii a and b (b > a) and proper length L' (very large compared to the radii). In S' the electric field is purely radial and there is no magnetic field.

	For a boost along z, the components of the fields transform according to:		For a boost along z, the components of the fields transform according to:

	$\begin{aligned}		$\begin{aligned}
	E_\parallel' &= E_\parallel, \quad B_\parallel' = B_\parallel,\\		E_\parallel' &= E_\parallel, \quad B_\parallel' = B_\parallel,\\
	\mathbf{E}_\perp' &= \gamma (\mathbf{E}_\perp + \mathbf{v} \times \mathbf{B}),\\		\mathbf{E}_\perp' &= \gamma (\mathbf{E}_\perp + \mathbf{v} \times \mathbf{B}),\\
	\mathbf{B}_\perp' &= \gamma \left(\mathbf{B}_\perp - \frac{\mathbf{v}}{c^2} \times \mathbf{E}\right).		\mathbf{B}_\perp' &= \gamma \left(\mathbf{B}_\perp - \frac{\mathbf{v}}{c^2} \times \mathbf{E}\right).
	\end{aligned}$		\end{aligned}$
	@@ -21,28 +23,28 @@Solution

	In our case, in S' (at rest):		In our case, in S' (at rest):

	~~$\mathbf{E}'~~ is radial (perpendicular to z), therefore \mathbf{E}_\perp' = \mathbf{E}', E_\parallel' = 0$		$\mathbf{E}'$
			is radial (perpendicular to z), therefore $\mathbf{E}_\perp' = \mathbf{E}', E_\parallel' = 0$
	$\mathbf{B}' = 0$		$\mathbf{B}' = 0$

	In S		In S
	using the inverse transformation (changing v \to -v)		using the inverse transformation (changing v \to -v) :
	:

	$\mathbf{E}_\perp = \gamma (\mathbf{E}_\perp' - \mathbf{v} \times \mathbf{B}') = \gamma \mathbf{E}_\perp', \quad \mathbf{E}_\parallel = 0$		$\mathbf{E}_\perp = \gamma (\mathbf{E}_\perp' - \mathbf{v} \times \mathbf{B}') = \gamma \mathbf{E}_\perp', \quad \mathbf{E}_\parallel = 0$

	$\mathbf{B}_\perp = \gamma \left(\mathbf{B}_\perp' + \frac{\mathbf{v}}{c^2} \times \mathbf{E}'\right) = \gamma \frac{\mathbf{v}}{c^2} \times \mathbf{E}'$.		$\mathbf{B}_\perp = \gamma \left(\mathbf{B}_\perp' + \frac{\mathbf{v}}{c^2} \times \mathbf{E}'\right) = \gamma \frac{\mathbf{v}}{c^2} \times \mathbf{E}'$

	The radial electric field in S is:		The radial electric field in S is:

	$\boxed{E_r = \gamma E_r'}$		$\boxed{E_r = \gamma E_r'}$

	Potential difference between the cylinders		Potential difference between the cylinders

	The potential difference in S' is $		The potential difference in S' is $
	V' = \int_a^b E_r' \, dr$.		V' = \int_a^b E_r' \, dr$
	In S:		In S:

	$V = \int_a^b E_r \, dr = \int_a^b \gamma E_r' \, dr = \gamma V'$.		$V = \int_a^b E_r \, dr = \int_a^b \gamma E_r' \, dr = \gamma V'$

	The transverse distances (a and b) do not change because the motion is parallel to the axis.		The transverse distances (a and b) do not change because the motion is parallel to the axis.

	$\boxed{V = \gamma V' = \frac{V'}{\sqrt{1-\beta^2}}}$.		$\boxed{V = \gamma V' = \frac{V'}{\sqrt{1-\beta^2}}}$.

	The potential difference increases by a factor of $ \gamma$		The potential difference increases by a factor of $ \gamma$

	Capacitance		Capacitance

	The total charge Q on the plates is a relativistic invariant (scalar). In S' the capacitance is $C' = Q/V'$		The total charge Q on the plates is a relativistic invariant (scalar). In S' the capacitance is $C' = Q/V'$
	In S:		In S:

	$C = \frac{Q}{V} = \frac{Q}{\gamma V'} = \frac{C'}{\gamma}$		$C = \frac{Q}{V} = \frac{Q}{\gamma V'} = \frac{C'}{\gamma}$

	$\boxed{C = C' \sqrt{1-\beta^2}}$		$\boxed{C = C' \sqrt{1-\beta^2}}$

	The capacitance decreases by a factor of \gamma (it is multiplied by		The capacitance decreases by a factor of \gamma (it is multiplied by
	$\sqrt{1-\beta^2})$		$\sqrt{1-\beta^2})$
	@@ -64,4 +66,6 @@Solution

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		$\boxed{C = C' \sqrt{1-\beta^2}}$

			$\boxed{V = \gamma V' = \frac{V'}{\sqrt{1-\beta^2}}}$.