1. **Problem:** Find the distance between charges $q_1 = +2\ \mu C$ and $q_2 = +35\ \mu C$ given that the electric field is zero at a point 16 cm from $q_2$. 2. **Formula:** The electric field $E$ due to a point charge $q$ at distance $r$ is given by $$E = \frac{k|q|}{r^2}$$ where $k = 9 \times 10^9\ \text{Nm}^2/\text{C}^2$. 3. **Key idea:** The net electric field is zero at a point where the fields from $q_1$ and $q_2$ cancel each other out. Since both charges are positive, the zero field point lies between them. 4. **Set distances:** Let the distance between $q_1$ and $q_2$ be $d$. The zero field point is 16 cm from $q_2$, so its distance from $q_1$ is $d - 0.16$ m. 5. **Set equation for zero field:** $$\frac{k q_1}{(d - 0.16)^2} = \frac{k q_2}{(0.16)^2}$$ Cancel $k$ and substitute values: $$\frac{2 \times 10^{-6}}{(d - 0.16)^2} = \frac{35 \times 10^{-6}}{(0.16)^2}$$ 6. **Solve for $d - 0.16$:}** $$\frac{2}{(d - 0.16)^2} = \frac{35}{0.0256}$$ $$\Rightarrow (d - 0.16)^2 = \frac{2 \times 0.0256}{35} = 0.0014629$$ $$d - 0.16 = \sqrt{0.0014629} = 0.03825\ \text{m}$$ 7. **Find $d$:}** $$d = 0.16 + 0.03825 = 0.19825\ \text{m} = 19.8\ \text{cm}$$ **Final answer:** The distance between $q_1$ and $q_2$ is approximately **19.8 cm**.