1. **State the problem:** We are given a charge $q = 6\ \mu C = 6 \times 10^{-6}\ C$ and a force $f = 2.4 \times 10^{-4}\ N$. We want to find the electric field $E$. 2. **Formula used:** The electric field $E$ is related to force $f$ and charge $q$ by the formula: $$E = \frac{f}{q}$$ 3. **Substitute the values:** $$E = \frac{2.4 \times 10^{-4}}{6 \times 10^{-6}}$$ 4. **Simplify the fraction:** $$E = \frac{2.4}{6} \times \frac{10^{-4}}{10^{-6}}$$ 5. **Calculate each part:** $$\frac{2.4}{6} = 0.4$$ $$\frac{10^{-4}}{10^{-6}} = 10^{(-4 - (-6))} = 10^{2} = 100$$ 6. **Multiply the results:** $$E = 0.4 \times 100 = 40$$ 7. **Interpretation:** The electric field $E$ is $40\ N/C$, and since the force is positive and the charge is positive, the electric field points away from the charge. **Final answer:** $$E = 40\ \text{N/C, going away from the charge}$$