Правка разделов «Solution», «Answer»
en/6.3.12.md
+5 −4
| ### Statement | |||
| $6.3.12.$ [Insert the problem statement] | |||
| ### Solution | |||
| @@ -6,10 +6,10 @@Solution | |||
| Let´s suppose the upper plate with density $+\sigma$ and the lower one with density $-\sigma$, each plate generates an electric field of modular value $E = \frac{\sigma}{2\varepsilon_0}$. Then, the net field between them is $E_n = \frac{\sigma}{\varepsilon_0}$ because vectors are summed up (they have the same direction). Finally,\ | |||
| $V = -\int_{r}^{0} \vec{E} \cdot d\vec{r}$\ | |||
| − | $V = \frac{\sigma}{\varepsilon_0}r$\ | ||
| − | Calculating, for $r$ = 1 cm, in CGS system. In this case, $\frac{1}{\varepsilon_0} = 4\pi$\ | ||
| + | $V(r) = \frac{\sigma}{\varepsilon_0}r$\ | ||
| + | a) Calculating, for $r$ = 1 cm, in CGS system. In this case, $\frac{1}{\varepsilon_0} = 4\pi$\ | ||
| $V = 4\pi \times 3\;\rm{\frac{esu}{cm^2}}\times 1\;\rm{cm}$\ | |||
| − | $V \simeq 37.7 | ||
| + | $V \simeq 37.7\;\rm{CGS}\;(\rm{statvolt})$\ | ||
| Taking in account that 1 statvolt = 299.792458 V,\ | |||
| $V \simeq 37.7\times 299.792458\;\rm{V} \simeq 11 300\;\rm{V} \simeq 11.3\;\rm{kV}$\ | |||
| b) For $r = 5$ cm, and 1 C = 2.9979$\times$10$^9$ esu,\ | |||
| $V =4\pi \times 10^{-10}\times 2,9979\times 10^9\;\rm{\frac{esu}{cm^2}}\times 5\;\rm{cm}$\ | |||
| $V = 18,85\;\rm{CGS}\;(\rm{statvolt})$\ | |||
| and in the SI,\ | |||
| $V = \frac{10^{-6}\;\rm{\frac{C}{m^2}}}{8.85\times10^{-12}\;\rm{\frac{C^2}{N\;m^2}}}\times 5\times10^{-2}\;\rm{m} \simeq 5650\;\rm{V} = 5.65\;\rm{kV}$ | |||
| @@ -25,4 +25,5 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | a) $V$ = 37.7 CGS, $V$ = 11.3 kV | ||
| + | b) $V$ = 18.85 CGS, $V$ = 5.65 kV | ||
| ### Statement | ### Statement | ||
| $6.3.12.$ [Insert the problem statement] | $6.3.12.$ [Insert the problem statement] | ||
| ### Solution | ### Solution | ||
| @@ -6,10 +6,10 @@Solution | |||
| Let´s suppose the upper plate with density $+\sigma$ and the lower one with density $-\sigma$, each plate generates an electric field of modular value $E = \frac{\sigma}{2\varepsilon_0}$. Then, the net field between them is $E_n = \frac{\sigma}{\varepsilon_0}$ because vectors are summed up (they have the same direction). Finally,\ | Let´s suppose the upper plate with density $+\sigma$ and the lower one with density $-\sigma$, each plate generates an electric field of modular value $E = \frac{\sigma}{2\varepsilon_0}$. Then, the net field between them is $E_n = \frac{\sigma}{\varepsilon_0}$ because vectors are summed up (they have the same direction). Finally,\ | ||
| $V = -\int_{r}^{0} \vec{E} \cdot d\vec{r}$\ | $V = -\int_{r}^{0} \vec{E} \cdot d\vec{r}$\ | ||
| $V = \frac{\sigma}{\varepsilon_0}r$\ | $V(r) = \frac{\sigma}{\varepsilon_0}r$\ | ||
| Calculating, for $r$ = 1 cm, in CGS system. In this case, $\frac{1}{\varepsilon_0} = 4\pi$\ | a) Calculating, for $r$ = 1 cm, in CGS system. In this case, $\frac{1}{\varepsilon_0} = 4\pi$\ | ||
| $V = 4\pi \times 3\;\rm{\frac{esu}{cm^2}}\times 1\;\rm{cm}$\ | $V = 4\pi \times 3\;\rm{\frac{esu}{cm^2}}\times 1\;\rm{cm}$\ | ||
| $V \simeq 37.7 |
$V \simeq 37.7\;\rm{CGS}\;(\rm{statvolt})$\ | ||
| Taking in account that 1 statvolt = 299.792458 V,\ | Taking in account that 1 statvolt = 299.792458 V,\ | ||
| $V \simeq 37.7\times 299.792458\;\rm{V} \simeq 11 300\;\rm{V} \simeq 11.3\;\rm{kV}$\ | $V \simeq 37.7\times 299.792458\;\rm{V} \simeq 11 300\;\rm{V} \simeq 11.3\;\rm{kV}$\ | ||
| b) For $r = 5$ cm, and 1 C = 2.9979$\times$10$^9$ esu,\ | b) For $r = 5$ cm, and 1 C = 2.9979$\times$10$^9$ esu,\ | ||
| $V =4\pi \times 10^{-10}\times 2,9979\times 10^9\;\rm{\frac{esu}{cm^2}}\times 5\;\rm{cm}$\ | $V =4\pi \times 10^{-10}\times 2,9979\times 10^9\;\rm{\frac{esu}{cm^2}}\times 5\;\rm{cm}$\ | ||
| $V = 18,85\;\rm{CGS}\;(\rm{statvolt})$\ | $V = 18,85\;\rm{CGS}\;(\rm{statvolt})$\ | ||
| and in the SI,\ | and in the SI,\ | ||
| $V = \frac{10^{-6}\;\rm{\frac{C}{m^2}}}{8.85\times10^{-12}\;\rm{\frac{C^2}{N\;m^2}}}\times 5\times10^{-2}\;\rm{m} \simeq 5650\;\rm{V} = 5.65\;\rm{kV}$ | $V = \frac{10^{-6}\;\rm{\frac{C}{m^2}}}{8.85\times10^{-12}\;\rm{\frac{C^2}{N\;m^2}}}\times 5\times10^{-2}\;\rm{m} \simeq 5650\;\rm{V} = 5.65\;\rm{kV}$ | ||
| @@ -25,4 +25,5 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | a) $V$ = 37.7 CGS, $V$ = 11.3 kV | ||
| b) $V$ = 18.85 CGS, $V$ = 5.65 kV | |||