Edits to “Statement”, “Answer”
en/6.1.1.md
+4 −2
| @@ -1,7 +1,8 @@ | |||
| ### Statement | |||
| − | $6.1.1.$ [a. Find the interaction force of charges of 1 and 2 C at a distance of 1 km from eachother, b. With what force do two electrons interact at a distance of 10^{-8} cm? How many times is this force greater than the force of their gravitational attraction?] | ||
| + | $6.1.1.$ {[a. Find the interaction force of charges of 1 and 2 C at a distance of 1 km from eachother, b. With what force do two electrons interact at a distance of $10^{-8}$ cm? How many times is this force greater than the force of their gravitational attraction?] | ||
| + | |||
| ### 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: | |||
| 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} \text{ N} | |||
| \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} | |||
| @@ -41,5 +42,6 @@Solution | |||
| #### Answer | |||
| − | |||
| + | $$ | ||
| [a) 1.8*10^4 N, b) 4.2*10^{42}] | |||
| + | $$ | ||
| @@ -1,7 +1,8 @@ | |||
| ### Statement | ### Statement | ||
| $6.1.1.$ [a. Find the interaction force of charges of 1 and 2 C at a distance of 1 km from eachother, b. With what force do two electrons interact at a distance of 10^{-8} cm? How many times is this force greater than the force of their gravitational attraction?] | $6.1.1.$ {[a. Find the interaction force of charges of 1 and 2 C at a distance of 1 km from eachother, b. With what force do two electrons interact at a distance of $10^{-8}$ cm? How many times is this force greater than the force of their gravitational attraction?] | ||
| ### 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: | ||
| 18,000 N or about $1.8\cdot 10^4$ N. | 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, | ||
| \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} | ||
| @@ -41,5 +42,6 @@Solution | |||
| #### Answer | #### Answer | ||
| $$ | |||
| [a) 1.8*10^4 N, b) 4.2*10^{42}] | [a) 1.8*10^4 N, b) 4.2*10^{42}] | ||
| $$ | |||