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### Statement
−$6.1.2.$ One cubic centimeter of water was divided into charges with opposite signs and then separated to a distance of 1 m. With what force will the charges attract each other?
+$6.1.2.$ The interaction force between two identical charges at a distance of $1\text{ m}$ is $1\text{ N}$. Determine these charges in SI and CGS systems.
### Solution
−![For problem $6.1.2$|2481x3506, 100%](../../img/6.1.2/Savchenko 6 1 2- JCFC.png)
+<b>1. In the SI system:</b>
+From Coulomb's law:
+$$F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q^2}{r^2}$$
+Expressing the charge $q$:
+$$q = \sqrt{4\pi\varepsilon_0 r^2 F}$$
+Substitute the numerical values ($F = 1\text{ N}$, $r = 1\text{ m}$, $\frac{1}{4\pi\varepsilon_0} = 9 \cdot 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$):
+$$q = \sqrt{\frac{1 \cdot 1^2}{9 \cdot 10^9}} = \frac{1}{3 \cdot 10^4 \cdot \sqrt{10}} \approx 1.05 \cdot 10^{-5}\text{ C}$$
−#### Answer
+<b>2. In the CGS system (electrostatic):</b>
+In the CGS system, the coefficient is $1$, and Coulomb's law is written as:
+$$F = \frac{q^2}{r^2}$$
+Expressing the charge $q_{\text{cgs}}$:
+$$q_{\text{cgs}} = r\sqrt{F}$$
+Keep in mind that in CGS, force is measured in dynes ($1\text{ N} = 10^5\text{ dyn}$), and distance is measured in centimeters ($1\text{ m} = 100\text{ cm}$).
+$$q_{\text{cgs}} = 100\text{ cm} \cdot \sqrt{10^5\text{ dyn}} = 100 \cdot 316.2 = 3.16 \cdot 10^4\text{ statC (esu)}$$
−$$
−F_e = \frac{\left(53{,}55 \times 10^3\right)^2}{4\pi \times 8{,}85 \times 10^{-12} \times 1^2} \quad \Rightarrow \quad \boxed{F_e = 2{,}58 \times 10^{19} \ \text{N}} \tag{7}
−$$
+#### Answer
+$1.05 \cdot 10^{-5}\text{ C}$; $3.16 \cdot 10^4\text{ statC}$