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