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en/6.1.1.md
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| + | ### Statement | ||
| + | |||
| + | $6.1.1.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | |||
| + | (a) By using Coulomb's Law, | ||
| + | \begin{equation} | ||
| + | F = \frac{1}{4\pi\epsilon_0}\cdot\frac{q_{1}q_{2}}{r^2} | ||
| + | \end{equation} | ||
| + | And plugging in the values given (1 C, 2C, 1 km/1000m), | ||
| + | \begin{equation} | ||
| + | F = \frac{1}{4\pi\epsilon_0}\cdot\frac{1\cdot2}{1000^2} = \frac{2}{10^{6}4\pi\epsilon_0} | ||
| + | \end{equation} | ||
| + | Calculating this value gives us: | ||
| + | \text{18,000 N or about $1.8\cdot 10^4$ N}. | ||
| + | |||
| + | (b) Once again, using Coulomb's Law: | ||
| + | \begin{equation} | ||
| + | F = \frac{1}{4\pi\epsilon_0} \cdot\frac{q_{1}q_{2}}{r^2} | ||
| + | \end{equation} | ||
| + | Knowing that the charge of an electron is $-1.6\cdot10^{-19}$, and their distance is $10^{-8}$ cm $= 10^{-10}$ m, | ||
| + | \begin{equation} | ||
| + | F = \frac{1}{4\pi\epsilon_0}\cdot\frac{({-1.6\cdot10^{-19}})^2}{({10^{-10}})^2} = \frac{2.56\cdot10^{-38}}{10^{-20}4\pi\epsilon_0} | ||
| + | \end{equation} | ||
| + | Computing this value gives $2.3\cdot10^{-8}$ N | ||
| + | If we compare this to the gravitational attraction of the electrons, which is first given by the Law of Universal Gravitation: | ||
| + | \begin{equation} | ||
| + | F = G\frac{m_{1}m_{2}}{r^2} | ||
| + | \end{equation} | ||
| + | and then plugging in the known distance and the electron mass of $9.1\cdot10^{-31}$ kg, | ||
| + | \begin{equation} | ||
| + | F = G\frac{({-9.1\cdot10^{-31}})^2}{({10^{-10}})^2} = 5.5\cdot10^{-51} | ||
| + | \end{equation} | ||
| + | Dividing these two values to give us the ratio between electromagnetic and gravitational attraction gives us: | ||
| + | \begin{equation} | ||
| + | \frac{2.3\cdot10^{-8}}{5.5\cdot10^{-51}} = 4.2\cdot10^{42} | ||
| + | \end{equation} | ||
| + | |||
| + | |||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $6.1.1.$ [Insert the problem statement] | |||
| ### Solution | |||
| (a) By using Coulomb's Law, | |||
| \begin{equation} | |||
| F = \frac{1}{4\pi\epsilon_0}\cdot\frac{q_{1}q_{2}}{r^2} | |||
| \end{equation} | |||
| And plugging in the values given (1 C, 2C, 1 km/1000m), | |||
| \begin{equation} | |||
| F = \frac{1}{4\pi\epsilon_0}\cdot\frac{1\cdot2}{1000^2} = \frac{2}{10^{6}4\pi\epsilon_0} | |||
| \end{equation} | |||
| Calculating this value gives us: | |||
| \text{18,000 N or about $1.8\cdot 10^4$ N}. | |||
| (b) Once again, using Coulomb's Law: | |||
| \begin{equation} | |||
| F = \frac{1}{4\pi\epsilon_0} \cdot\frac{q_{1}q_{2}}{r^2} | |||
| \end{equation} | |||
| Knowing that the charge of an electron is $-1.6\cdot10^{-19}$, and their distance is $10^{-8}$ cm $= 10^{-10}$ m, | |||
| \begin{equation} | |||
| F = \frac{1}{4\pi\epsilon_0}\cdot\frac{({-1.6\cdot10^{-19}})^2}{({10^{-10}})^2} = \frac{2.56\cdot10^{-38}}{10^{-20}4\pi\epsilon_0} | |||
| \end{equation} | |||
| Computing this value gives $2.3\cdot10^{-8}$ N | |||
| If we compare this to the gravitational attraction of the electrons, which is first given by the Law of Universal Gravitation: | |||
| \begin{equation} | |||
| F = G\frac{m_{1}m_{2}}{r^2} | |||
| \end{equation} | |||
| and then plugging in the known distance and the electron mass of $9.1\cdot10^{-31}$ kg, | |||
| \begin{equation} | |||
| F = G\frac{({-9.1\cdot10^{-31}})^2}{({10^{-10}})^2} = 5.5\cdot10^{-51} | |||
| \end{equation} | |||
| Dividing these two values to give us the ratio between electromagnetic and gravitational attraction gives us: | |||
| \begin{equation} | |||
| \frac{2.3\cdot10^{-8}}{5.5\cdot10^{-51}} = 4.2\cdot10^{42} | |||
| \end{equation} | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||