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+### 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.
+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:
+$$\frac{F_e}{F_g} = \frac{2.3 \cdot 10^{-8}}{5.5 \cdot 10^{-51}} = 4.2 \cdot 10^{42}$$
+
+#### Answer
+a) $1.8 \cdot 10^4\text{ N}$
+b) $4.2 \cdot 10^{42}$ times
+
+#### Answer
+
+[Insert a concise answer or boxed result]