1. **State the problem:** We have a force $F = 5$ newtons acting on a charge $q = 6$ microcoulombs at a point. We need to find the electric field $E$ and intensity at that point. 2. **Formula used:** The electric field $E$ at a point due to a force on a charge is given by: $$E = \frac{F}{q}$$ where $E$ is the electric field in newtons per coulomb (N/C), $F$ is the force in newtons (N), and $q$ is the charge in coulombs (C). 3. **Important rules:** - Convert microcoulombs to coulombs: $1$ microcoulomb = $10^{-6}$ coulombs. - The electric field direction is the direction of the force on a positive charge. 4. **Convert charge:** $$q = 6 \text{ microcoulombs} = 6 \times 10^{-6} \text{ C}$$ 5. **Calculate electric field:** $$E = \frac{F}{q} = \frac{5}{6 \times 10^{-6}}$$ 6. **Simplify the expression:** $$E = \frac{5}{6 \times 10^{-6}} = \frac{5}{\cancel{6} \times 10^{-6}} \times \frac{1}{\cancel{6}} = \frac{5}{6} \times 10^{6} = 0.8333 \times 10^{6} = 8.33 \times 10^{5}$$ 7. **Final answer:** The electric field intensity at that point is: $$E = 8.33 \times 10^{5} \text{ N/C}$$ This means the electric field strength is $8.33 \times 10^{5}$ newtons per coulomb at that point, directed along the force acting on the positive charge.