Правка разделов «Statement», «Answer»
en/6.1.3.md
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| ### Statement | |||
| − | $6.1.3.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
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| − | ### Statement | ||
| − | |||
| $6.1.3.$ | |||
| − | a. The force acting on a charge of $1\text{ C}$ is $1\text{ N}$. What is the electric field strength acting on this charge, in SI and CGS? | ||
| + | a. The force acting on a charge of $1\text{ C}$ is $1\text{ N}$. What is the electric field strength acting on this charge, in SI and CGS?\ | ||
| b. The force acting on a charge of $10\text{ CGS units (statC)}$ is $100\text{ dyn}$. What is the electric field strength acting on this charge, in SI and CGS? | |||
| ### Solution | |||
| <b>a)</b> According to the definition of electric field strength: | |||
| $$F = qE \implies E = \frac{F}{q}$$ | |||
| In the SI system: | |||
| $$E = \frac{1\text{ N}}{1\text{ C}} = 1\text{ V/m}$$ | |||
| To convert to CGS, we use the relations: $1\text{ V} = \frac{1}{300}\text{ statV}$ and $1\text{ m} = 100\text{ cm}$: | |||
| $$E = \frac{1/300\text{ statV}}{100\text{ cm}} = \frac{1}{30000}\text{ statV/cm} \approx 3.33 \cdot 10^{-5}\text{ statV/cm}$$ | |||
| <b>b)</b> For the second case, we first calculate the field strength in the CGS system: | |||
| $$E_{\text{cgs}} = \frac{F}{q} = \frac{100\text{ dyn}}{10\text{ statC}} = 10\text{ statV/cm}$$ | |||
| Now we convert this value to the SI system using the inverse relations ($1\text{ statV} = 300\text{ V}$, $1\text{ cm} = 0.01\text{ m}$): | |||
| $$E = 10 \cdot \frac{300\text{ V}}{0.01\text{ m}} = 10 \cdot 30000\text{ V/m} = 3 \cdot 10^5\text{ V/m}$$ | |||
| @@ -27,7 +21,3 @@Solution | |||
| #### Answer | |||
| a. $1\text{ V/m} \approx 3.33 \cdot 10^{-5}\text{ statV/cm}$ | |||
| b. $10\text{ statV/cm} = 3 \cdot 10^5\text{ V/m}$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,13 +1,7 @@ | |||
| ### Statement | ### Statement | ||
| $6.1.3.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $6.1.3.$ | $6.1.3.$ | ||
| a. The force acting on a charge of $1\text{ C}$ is $1\text{ N}$. What is the electric field strength acting on this charge, in SI and CGS? | a. The force acting on a charge of $1\text{ C}$ is $1\text{ N}$. What is the electric field strength acting on this charge, in SI and CGS?\ | ||
| b. The force acting on a charge of $10\text{ CGS units (statC)}$ is $100\text{ dyn}$. What is the electric field strength acting on this charge, in SI and CGS? | b. The force acting on a charge of $10\text{ CGS units (statC)}$ is $100\text{ dyn}$. What is the electric field strength acting on this charge, in SI and CGS? | ||
| ### Solution | ### Solution | ||
| <b>a)</b> According to the definition of electric field strength: | <b>a)</b> According to the definition of electric field strength: | ||
| $$F = qE \implies E = \frac{F}{q}$$ | $$F = qE \implies E = \frac{F}{q}$$ | ||
| In the SI system: | In the SI system: | ||
| $$E = \frac{1\text{ N}}{1\text{ C}} = 1\text{ V/m}$$ | $$E = \frac{1\text{ N}}{1\text{ C}} = 1\text{ V/m}$$ | ||
| To convert to CGS, we use the relations: $1\text{ V} = \frac{1}{300}\text{ statV}$ and $1\text{ m} = 100\text{ cm}$: | To convert to CGS, we use the relations: $1\text{ V} = \frac{1}{300}\text{ statV}$ and $1\text{ m} = 100\text{ cm}$: | ||
| $$E = \frac{1/300\text{ statV}}{100\text{ cm}} = \frac{1}{30000}\text{ statV/cm} \approx 3.33 \cdot 10^{-5}\text{ statV/cm}$$ | $$E = \frac{1/300\text{ statV}}{100\text{ cm}} = \frac{1}{30000}\text{ statV/cm} \approx 3.33 \cdot 10^{-5}\text{ statV/cm}$$ | ||
| <b>b)</b> For the second case, we first calculate the field strength in the CGS system: | <b>b)</b> For the second case, we first calculate the field strength in the CGS system: | ||
| $$E_{\text{cgs}} = \frac{F}{q} = \frac{100\text{ dyn}}{10\text{ statC}} = 10\text{ statV/cm}$$ | $$E_{\text{cgs}} = \frac{F}{q} = \frac{100\text{ dyn}}{10\text{ statC}} = 10\text{ statV/cm}$$ | ||
| Now we convert this value to the SI system using the inverse relations ($1\text{ statV} = 300\text{ V}$, $1\text{ cm} = 0.01\text{ m}$): | Now we convert this value to the SI system using the inverse relations ($1\text{ statV} = 300\text{ V}$, $1\text{ cm} = 0.01\text{ m}$): | ||
| $$E = 10 \cdot \frac{300\text{ V}}{0.01\text{ m}} = 10 \cdot 30000\text{ V/m} = 3 \cdot 10^5\text{ V/m}$$ | $$E = 10 \cdot \frac{300\text{ V}}{0.01\text{ m}} = 10 \cdot 30000\text{ V/m} = 3 \cdot 10^5\text{ V/m}$$ | ||
| @@ -27,7 +21,3 @@Solution | |||
| #### Answer | #### Answer | ||
| a. $1\text{ V/m} \approx 3.33 \cdot 10^{-5}\text{ statV/cm}$ | a. $1\text{ V/m} \approx 3.33 \cdot 10^{-5}\text{ statV/cm}$ | ||
| b. $10\text{ statV/cm} = 3 \cdot 10^5\text{ V/m}$ | b. $10\text{ statV/cm} = 3 \cdot 10^5\text{ V/m}$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||