1. **State the problem:** We have two point charges $Q_1 = Q_2 = 2 \times 10^{-12}$ C separated by a distance $a = 3 \times 10^{-2}$ m. We want to find the electric field $E_A$ at point $A$ (midpoint between the charges) and the force $F$ between the charges. 2. **Calculate the force $F$ between the charges:** Using Coulomb's law: $$F = k \frac{|Q_1 Q_2|}{a^2}$$ where $k = 8.99 \times 10^9$ N m$^2$/C$^2$. Substitute values: $$F = 8.99 \times 10^9 \times \frac{(2 \times 10^{-12})^2}{(3 \times 10^{-2})^2}$$ Calculate numerator: $$(2 \times 10^{-12})^2 = 4 \times 10^{-24}$$ Calculate denominator: $$(3 \times 10^{-2})^2 = 9 \times 10^{-4}$$ So, $$F = 8.99 \times 10^9 \times \frac{4 \times 10^{-24}}{9 \times 10^{-4}} = 8.99 \times 10^9 \times 4.44 \times 10^{-21} = 3.99 \times 10^{-11} \text{ N}$$ 3. **Calculate the electric field $E_A$ at midpoint $A$:** The electric field due to a point charge is: $$E = k \frac{|Q|}{r^2}$$ At midpoint $A$, distance from each charge is $r = \frac{a}{2} = 1.5 \times 10^{-2}$ m. Calculate magnitude of electric field from one charge at $A$: $$E_1 = 8.99 \times 10^9 \times \frac{2 \times 10^{-12}}{(1.5 \times 10^{-2})^2}$$ Calculate denominator: $$(1.5 \times 10^{-2})^2 = 2.25 \times 10^{-4}$$ So, $$E_1 = 8.99 \times 10^9 \times \frac{2 \times 10^{-12}}{2.25 \times 10^{-4}} = 8.99 \times 10^9 \times 8.89 \times 10^{-9} = 79.9 \text{ N/C}$$ 4. **Determine direction and net electric field $E_A$:** The two fields $E_{1A}$ and $E_{2A}$ from charges $Q_1$ and $Q_2$ at point $A$ are equal in magnitude but point upward and slightly to the right and left respectively. Their horizontal components cancel out, and vertical components add. Vertical component of one field: $$E_{1y} = E_1 \cos(\theta)$$ where $\theta$ is the angle between the line connecting charge to $A$ and vertical. Since charges are horizontally separated by $a$, and $A$ is midpoint, $\theta = 45^\circ$. Calculate vertical component: $$E_{1y} = 79.9 \times \cos 45^\circ = 79.9 \times \frac{\sqrt{2}}{2} = 56.5 \text{ N/C}$$ Total vertical field: $$E_A = 2 \times E_{1y} = 2 \times 56.5 = 113 \text{ N/C}$$ **Final answers:** $$F = 3.99 \times 10^{-11} \text{ N}$$ $$E_A = 113 \text{ N/C}$$