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Alexphysics edited Jun 17, 2026, 6:07 AM

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		+	### Statement
		+
		+	$14.3.11.$ [Insert the problem statement]
		+
		+	### 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).
		+
		+	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:
		+
		+	$\begin{aligned}
		+	E_\parallel' &= E_\parallel, \quad B_\parallel' = B_\parallel,\\
		+	\mathbf{E}_\perp' &= \gamma (\mathbf{E}_\perp + \mathbf{v} \times \mathbf{B}),\\
		+	\mathbf{B}_\perp' &= \gamma \left(\mathbf{B}_\perp - \frac{\mathbf{v}}{c^2} \times \mathbf{E}\right).
		+	\end{aligned}$
		+
		+	In our case, in S' (at rest):
		+
		+	$\mathbf{E}' is radial (perpendicular to z), therefore \mathbf{E}_\perp' = \mathbf{E}', E_\parallel' = 0$
		+	$\mathbf{B}' = 0$
		+
		+	In S
		+	using the inverse transformation (changing v \to -v)
		+	:
		+
		+	$\mathbf{E}_\perp = \gamma (\mathbf{E}_\perp' - \mathbf{v} \times \mathbf{B}') = \gamma \mathbf{E}_\perp', \quad \mathbf{E}_\parallel = 0$
		+
		+	$\mathbf{B}_\perp = \gamma \left(\mathbf{B}_\perp' + \frac{\mathbf{v}}{c^2} \times \mathbf{E}'\right) = \gamma \frac{\mathbf{v}}{c^2} \times \mathbf{E}'$.
		+
		+	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$.
		+	In S:
		+
		+	$V = \int_a^b E_r \, dr = \int_a^b \gamma E_r' \, dr = \gamma V'$.
		+
		+	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
		+	$\sqrt{1-\beta^2})$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$14.3.11.$ [Insert the problem statement]

			### 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).

			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:

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

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

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

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

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

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

			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$.
			In S:

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

			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
			$\sqrt{1-\beta^2})$

			#### Answer

			[Insert a concise answer or boxed result]