New solution

Alexphysics edited Jun 15, 2026, 3:12 AM

View solution History

revision #18788 newer →

en/6.3.34.md +47 −0

		@@ -0,0 +1,47 @@
		+	### Statement
		+
		+	$6.3.34.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	The error here is ignoring the conservative nature of the electrostatic field, in addition to the symmetry of the ring, of course.
		+
		+	The incorrect assumption is that:
		+
		+	1. The electric field is strictly confined between the plates and is zero outside them.
		+	2. The net force on the portion of the ring located between the plates produces an unbalanced torque.
		+
		+	What actually happens is that, although the field is concentrated between the plates, the field lines at the edges curve and fields exist outside the interplate region.
		+	These fields then act on the rest of the ring and generate forces that exactly cancel any net torque.
		+
		+	Furthermore, since the electrostatic field is conservative, the potential energy of the ring cannot change over a complete rotation, so no net work can be extracted cyclically.
		+
		+	Consider a rigid ring with uniform linear charge density \lambda > 0, radius a, which can rotate freely about its axis. It is immersed in an arbitrary external electrostatic field \mathbf{E}(\mathbf{r}) = -\nabla V(\mathbf{r}).
		+
		+	We want to show that the resultant torque of the electric forces about the center of the ring, \mathbf{M}, is identically zero.
		+	The electrostatic potential energy of the ring in the external field is:
		+
		+	U = \int_{\text{ring}} V(\mathbf{r})\, dq = \lambda \oint V(\mathbf{r})\, dl
		+
		+	where dq = \lambda dl and the integral is taken over the circumference of the ring.
		+
		+	If we rotate the ring by an angle \theta about its axis, the charge distribution in space changes because the ring is a rigid object. The new potential energy will be a function of \theta: U(\theta). From mechanics, the total torque of the forces on a rigid system that can rotate is:
		+
		+	M = -\frac{dU}{d\theta}
		+
		+	(where M is the component of the torque along the rotation axis, with the appropriate sign).
		+
		+	Now, the charge distribution of the ring is invariant under rotation: when the ring is rotated by an arbitrary angle, the spatial charge distribution is exactly the same as at the beginning (because the charge is uniformly distributed around the ring).
		+	Since the external field \mathbf{E}(\mathbf{r}) is fixed (does not change with time), the potential energy U does not depend on \theta:
		+
		+	U(\theta) = \text{constant} \quad \Rightarrow \quad \frac{dU}{d\theta} = 0
		+
		+	Therefore,
		+
		+	{M = 0}
		+
		+	There is no net torque to accelerate the rotation. Any initial motion would be damped by losses (friction, air resistance) until it stops.
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,47 @@
			### Statement

			$6.3.34.$ [Insert the problem statement]

			### Solution

			The error here is ignoring the conservative nature of the electrostatic field, in addition to the symmetry of the ring, of course.

			The incorrect assumption is that:

			1. The electric field is strictly confined between the plates and is zero outside them.
			2. The net force on the portion of the ring located between the plates produces an unbalanced torque.

			What actually happens is that, although the field is concentrated between the plates, the field lines at the edges curve and fields exist outside the interplate region.
			These fields then act on the rest of the ring and generate forces that exactly cancel any net torque.

			Furthermore, since the electrostatic field is conservative, the potential energy of the ring cannot change over a complete rotation, so no net work can be extracted cyclically.

			Consider a rigid ring with uniform linear charge density \lambda > 0, radius a, which can rotate freely about its axis. It is immersed in an arbitrary external electrostatic field \mathbf{E}(\mathbf{r}) = -\nabla V(\mathbf{r}).

			We want to show that the resultant torque of the electric forces about the center of the ring, \mathbf{M}, is identically zero.
			The electrostatic potential energy of the ring in the external field is:

			U = \int_{\text{ring}} V(\mathbf{r})\, dq = \lambda \oint V(\mathbf{r})\, dl

			where dq = \lambda dl and the integral is taken over the circumference of the ring.

			If we rotate the ring by an angle \theta about its axis, the charge distribution in space changes because the ring is a rigid object. The new potential energy will be a function of \theta: U(\theta). From mechanics, the total torque of the forces on a rigid system that can rotate is:

			M = -\frac{dU}{d\theta}

			(where M is the component of the torque along the rotation axis, with the appropriate sign).

			Now, the charge distribution of the ring is invariant under rotation: when the ring is rotated by an arbitrary angle, the spatial charge distribution is exactly the same as at the beginning (because the charge is uniformly distributed around the ring).
			Since the external field \mathbf{E}(\mathbf{r}) is fixed (does not change with time), the potential energy U does not depend on \theta:

			U(\theta) = \text{constant} \quad \Rightarrow \quad \frac{dU}{d\theta} = 0

			Therefore,

			{M = 0}

			There is no net torque to accelerate the rotation. Any initial motion would be damped by losses (friction, air resistance) until it stops.

			#### Answer

			[Insert a concise answer or boxed result]