Edits to “Statement”, “Answer”
en/6.1.1.md
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| @@ -1,13 +1,7 @@ | |||
| ### Statement | |||
| − | $6.1.1.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| $6.1.1.$ | |||
| − | a. Find the interaction force of charges of $1$ and $2\text{ C}$ at a distance of $1\text{ km}$ from each other. | ||
| + | a. Find the interaction force of charges of $1$ and $2\text{ C}$ at a distance of $1\text{ km}$ from each other.\ | ||
| b. With what force do two electrons interact at a distance of $10^{-8}\text{ cm}$? How many times is this force greater than the force of their gravitational attraction? | |||
| ### Solution | |||
| <b>a)</b> By using Coulomb's Law: | |||
| $$F = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1 q_2}{r^2}$$ | |||
| Plugging in the given values ($1\text{ C}$, $2\text{ C}$, $1\text{ km} = 1000\text{ m}$): | |||
| $$F = \frac{1}{4\pi\varepsilon_0}\cdot\frac{1\cdot 2}{1000^2} = \frac{2}{10^6 \cdot 4\pi\varepsilon_0}$$ | |||
| Substituting the constant $\frac{1}{4\pi\varepsilon_0} = 9 \cdot 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$, calculating this value gives us: | |||
| $$F = 2 \cdot 10^{-6} \cdot 9 \cdot 10^9 = 18,000\text{ N} = 1.8 \cdot 10^4\text{ N}$$ | |||
| <b>b)</b> Once again, using Coulomb's Law: | |||
| $$F_e = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1 q_2}{r^2}$$ | |||
| Knowing that the charge of an electron is $e = -1.6 \cdot 10^{-19}\text{ C}$, and their distance is $10^{-8}\text{ cm} = 10^{-10}\text{ m}$: | |||
| $$F_e = \frac{1}{4\pi\varepsilon_0}\cdot\frac{(-1.6 \cdot 10^{-19})^2}{(10^{-10})^2} = 9 \cdot 10^9 \cdot \frac{2.56 \cdot 10^{-38}}{10^{-20}} = 2.3 \cdot 10^{-8}\text{ N}$$ | |||
| Now we compare this to the gravitational attraction of the electrons, given by the Law of Universal Gravitation: | |||
| $$F_g = G\frac{m_1 m_2}{r^2}$$ | |||
| Plugging in the known distance and the electron mass of $m = 9.1 \cdot 10^{-31}\text{ kg}$: | |||
| $$F_g = 6.67 \cdot 10^{-11} \cdot \frac{(9.1 \cdot 10^{-31})^2}{(10^{-10})^2} = 5.5 \cdot 10^{-51}\text{ N}$$ | |||
| Dividing these two values to find the ratio between electromagnetic and gravitational attraction gives us: | |||
| $$\frac{F_e}{F_g} = \frac{2.3 \cdot 10^{-8}}{5.5 \cdot 10^{-51}} = 4.2 \cdot 10^{42}$$ | |||
| @@ -33,7 +27,3 @@Solution | |||
| #### Answer | |||
| a) $1.8 \cdot 10^4\text{ N}$ | |||
| b) $4.2 \cdot 10^{42}$ times | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,13 +1,7 @@ | |||
| ### Statement | ### Statement | ||
| $6.1.1.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $6.1.1.$ | $6.1.1.$ | ||
| a. Find the interaction force of charges of $1$ and $2\text{ C}$ at a distance of $1\text{ km}$ from each other. | a. Find the interaction force of charges of $1$ and $2\text{ C}$ at a distance of $1\text{ km}$ from each other.\ | ||
| b. With what force do two electrons interact at a distance of $10^{-8}\text{ cm}$? How many times is this force greater than the force of their gravitational attraction? | b. With what force do two electrons interact at a distance of $10^{-8}\text{ cm}$? How many times is this force greater than the force of their gravitational attraction? | ||
| ### Solution | ### Solution | ||
| <b>a)</b> By using Coulomb's Law: | <b>a)</b> By using Coulomb's Law: | ||
| $$F = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1 q_2}{r^2}$$ | $$F = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1 q_2}{r^2}$$ | ||
| Plugging in the given values ($1\text{ C}$, $2\text{ C}$, $1\text{ km} = 1000\text{ m}$): | Plugging in the given values ($1\text{ C}$, $2\text{ C}$, $1\text{ km} = 1000\text{ m}$): | ||
| $$F = \frac{1}{4\pi\varepsilon_0}\cdot\frac{1\cdot 2}{1000^2} = \frac{2}{10^6 \cdot 4\pi\varepsilon_0}$$ | $$F = \frac{1}{4\pi\varepsilon_0}\cdot\frac{1\cdot 2}{1000^2} = \frac{2}{10^6 \cdot 4\pi\varepsilon_0}$$ | ||
| Substituting the constant $\frac{1}{4\pi\varepsilon_0} = 9 \cdot 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$, calculating this value gives us: | Substituting the constant $\frac{1}{4\pi\varepsilon_0} = 9 \cdot 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$, calculating this value gives us: | ||
| $$F = 2 \cdot 10^{-6} \cdot 9 \cdot 10^9 = 18,000\text{ N} = 1.8 \cdot 10^4\text{ N}$$ | $$F = 2 \cdot 10^{-6} \cdot 9 \cdot 10^9 = 18,000\text{ N} = 1.8 \cdot 10^4\text{ N}$$ | ||
| <b>b)</b> Once again, using Coulomb's Law: | <b>b)</b> Once again, using Coulomb's Law: | ||
| $$F_e = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1 q_2}{r^2}$$ | $$F_e = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1 q_2}{r^2}$$ | ||
| Knowing that the charge of an electron is $e = -1.6 \cdot 10^{-19}\text{ C}$, and their distance is $10^{-8}\text{ cm} = 10^{-10}\text{ m}$: | Knowing that the charge of an electron is $e = -1.6 \cdot 10^{-19}\text{ C}$, and their distance is $10^{-8}\text{ cm} = 10^{-10}\text{ m}$: | ||
| $$F_e = \frac{1}{4\pi\varepsilon_0}\cdot\frac{(-1.6 \cdot 10^{-19})^2}{(10^{-10})^2} = 9 \cdot 10^9 \cdot \frac{2.56 \cdot 10^{-38}}{10^{-20}} = 2.3 \cdot 10^{-8}\text{ N}$$ | $$F_e = \frac{1}{4\pi\varepsilon_0}\cdot\frac{(-1.6 \cdot 10^{-19})^2}{(10^{-10})^2} = 9 \cdot 10^9 \cdot \frac{2.56 \cdot 10^{-38}}{10^{-20}} = 2.3 \cdot 10^{-8}\text{ N}$$ | ||
| Now we compare this to the gravitational attraction of the electrons, given by the Law of Universal Gravitation: | Now we compare this to the gravitational attraction of the electrons, given by the Law of Universal Gravitation: | ||
| $$F_g = G\frac{m_1 m_2}{r^2}$$ | $$F_g = G\frac{m_1 m_2}{r^2}$$ | ||
| Plugging in the known distance and the electron mass of $m = 9.1 \cdot 10^{-31}\text{ kg}$: | Plugging in the known distance and the electron mass of $m = 9.1 \cdot 10^{-31}\text{ kg}$: | ||
| $$F_g = 6.67 \cdot 10^{-11} \cdot \frac{(9.1 \cdot 10^{-31})^2}{(10^{-10})^2} = 5.5 \cdot 10^{-51}\text{ N}$$ | $$F_g = 6.67 \cdot 10^{-11} \cdot \frac{(9.1 \cdot 10^{-31})^2}{(10^{-10})^2} = 5.5 \cdot 10^{-51}\text{ N}$$ | ||
| Dividing these two values to find the ratio between electromagnetic and gravitational attraction gives us: | Dividing these two values to find the ratio between electromagnetic and gravitational attraction gives us: | ||
| $$\frac{F_e}{F_g} = \frac{2.3 \cdot 10^{-8}}{5.5 \cdot 10^{-51}} = 4.2 \cdot 10^{42}$$ | $$\frac{F_e}{F_g} = \frac{2.3 \cdot 10^{-8}}{5.5 \cdot 10^{-51}} = 4.2 \cdot 10^{42}$$ | ||
| @@ -33,7 +27,3 @@Solution | |||
| #### Answer | #### Answer | ||
| a) $1.8 \cdot 10^4\text{ N}$ | a) $1.8 \cdot 10^4\text{ N}$ | ||
| b) $4.2 \cdot 10^{42}$ times | b) $4.2 \cdot 10^{42}$ times | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||