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