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+### Statement
+
+$6.1.3.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$6.1.3.$
+a. The force acting on a charge of $1\text{ C}$ is $1\text{ N}$. What is the electric field strength acting on this charge, in SI and CGS?
+b. The force acting on a charge of $10\text{ CGS units (statC)}$ is $100\text{ dyn}$. What is the electric field strength acting on this charge, in SI and CGS?
+
+### Solution
+
+<b>a)</b> According to the definition of electric field strength:
+$$F = qE \implies E = \frac{F}{q}$$
+In the SI system:
+$$E = \frac{1\text{ N}}{1\text{ C}} = 1\text{ V/m}$$
+To convert to CGS, we use the relations: $1\text{ V} = \frac{1}{300}\text{ statV}$ and $1\text{ m} = 100\text{ cm}$:
+$$E = \frac{1/300\text{ statV}}{100\text{ cm}} = \frac{1}{30000}\text{ statV/cm} \approx 3.33 \cdot 10^{-5}\text{ statV/cm}$$
+
+<b>b)</b> For the second case, we first calculate the field strength in the CGS system:
+$$E_{\text{cgs}} = \frac{F}{q} = \frac{100\text{ dyn}}{10\text{ statC}} = 10\text{ statV/cm}$$
+Now we convert this value to the SI system using the inverse relations ($1\text{ statV} = 300\text{ V}$, $1\text{ cm} = 0.01\text{ m}$):
+$$E = 10 \cdot \frac{300\text{ V}}{0.01\text{ m}} = 10 \cdot 30000\text{ V/m} = 3 \cdot 10^5\text{ V/m}$$
+
+#### Answer
+a. $1\text{ V/m} \approx 3.33 \cdot 10^{-5}\text{ statV/cm}$
+b. $10\text{ statV/cm} = 3 \cdot 10^5\text{ V/m}$
+
+#### Answer
+
+[Insert a concise answer or boxed result]