| ### Statement | | ### Statement |
| | | |
| $6.1.1.$ [Insert the problem statement] | | $6.1.1.$ [Insert the problem statement] |
| | | |
| ### Solution | | ### Solution |
| | | |
| | | |
| (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}. |
| | | |
| (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, |
| @@ -24,14 +24,14 @@Solution |
| \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} |
| 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} |
