Правка разделов «Condition», «Solution», «Answer»
en/6.5.7.md
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| @@ -1,11 +1,34 @@ | |||
| − | ### | ||
| + | ### Condition | ||
| + | $6.5.7.$ What charge can be placed per unit length of a long cylindrical shell of radius $R$, if, when the gas inside it is pumped, it withstands a pressure $P$? | ||
| − | $6.5.7.$ [Insert the problem statement] | ||
| − | |||
| ### Solution | |||
| + | Essentially, we need to find the charge at which the electric field pressure becomes equal to $P$. | ||
| − | 1 | ||
| + | The surface charge density is | ||
| + | $$ | ||
| + | \sigma = \frac{\rho}{2\pi R} \tag{1} | ||
| + | $$ | ||
| + | where $\rho = \frac{dq}{dl}$ is the desired quantity. | ||
| − | #### Answer | ||
| + | Near the surface of the cylinder, the field is | ||
| + | $$ | ||
| + | E = \frac{\sigma}{\varepsilon_0} \tag{2} | ||
| + | $$ | ||
| + | This is trivially derived from Gauss's theorem. | ||
| − | [Insert a concise answer or boxed result] | ||
| + | In the previous problem, the formula for the field pressure was proved (it is, incidentally, equal to the energy density of the field): | ||
| + | $$ | ||
| + | P = \frac{\varepsilon_0 E^2}{2} \tag{3} | ||
| + | $$ | ||
| + | $(1)\to(2)\to(3):$ | ||
| + | $$ | ||
| + | P = \frac{\rho^2}{8\varepsilon_0 \pi^2 R^2} \tag{3} | ||
| + | $$ | ||
| + | $$ | ||
| + | \rho = 2\pi R \sqrt{2\varepsilon_0 P} | ||
| + | $$ | ||
| + | |||
| + | #### Answer | ||
| + | $$ | ||
| + | \boxed{\rho = 2\pi R \sqrt{2\varepsilon_0 P}} | ||
| + | $$ | ||
| @@ -1,11 +1,34 @@ | |||
| ### |
### Condition | ||
| $6.5.7.$ What charge can be placed per unit length of a long cylindrical shell of radius $R$, if, when the gas inside it is pumped, it withstands a pressure $P$? | |||
| $6.5.7.$ [Insert the problem statement] | |||
| ### Solution | ### Solution | ||
| Essentially, we need to find the charge at which the electric field pressure becomes equal to $P$. | |||
| 1 | The surface charge density is | ||
| $$ | |||
| \sigma = \frac{\rho}{2\pi R} \tag{1} | |||
| $$ | |||
| where $\rho = \frac{dq}{dl}$ is the desired quantity. | |||
| #### Answer | Near the surface of the cylinder, the field is | ||
| $$ | |||
| E = \frac{\sigma}{\varepsilon_0} \tag{2} | |||
| $$ | |||
| This is trivially derived from Gauss's theorem. | |||
| [Insert a concise answer or boxed result] | In the previous problem, the formula for the field pressure was proved (it is, incidentally, equal to the energy density of the field): | ||
| $$ | |||
| P = \frac{\varepsilon_0 E^2}{2} \tag{3} | |||
| $$ | |||
| $(1)\to(2)\to(3):$ | |||
| $$ | |||
| P = \frac{\rho^2}{8\varepsilon_0 \pi^2 R^2} \tag{3} | |||
| $$ | |||
| $$ | |||
| \rho = 2\pi R \sqrt{2\varepsilon_0 P} | |||
| $$ | |||
| #### Answer | |||
| $$ | |||
| \boxed{\rho = 2\pi R \sqrt{2\varepsilon_0 P}} | |||
| $$ | |||