Новое решение
en/6.3.34.md
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| + | ### 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] | ||
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| ### 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] | |||