1. **State the problem:** We have a force $F = 100$ N acting north on a point charge $q = -20$ microcoulombs ($-20 \times 10^{-6}$ C). We need to find the magnitude and direction of the electric field $E$ at this point. 2. **Formula used:** The electric field $E$ at a point where a charge $q$ experiences a force $F$ is given by: $$E = \frac{F}{|q|}$$ 3. **Important rules:** - The electric field direction is the direction of the force on a positive test charge. - Since the charge here is negative, the force direction is opposite to the electric field direction. 4. **Calculate magnitude:** $$E = \frac{100}{20 \times 10^{-6}} = \frac{100}{0.00002} = 5,000,000 \text{ N/C}$$ 5. **Determine direction:** - Force is directed north on a negative charge. - Electric field direction is opposite to force for negative charge. - Therefore, electric field points south. **Final answer:** The magnitude of the electric field is $5,000,000$ N/C and its direction is south.