1. **Problem statement:** We have two vertical parallel lines, each of length $L=6.00\ \text{cm}=0.06\ \text{m}$, separated by a distance $d=6.00\ \text{cm}=0.06\ \text{m}$. They carry opposite linear charge densities $\lambda=1.00\ \mu\text{C/m}=1.00\times10^{-6}\ \text{C/m}$. We want to find the net electric field magnitude and direction at a point midway between the lines and $40.0\ \text{cm}=0.40\ \text{m}$ above their plane. 2. **Formula and concepts:** The electric field $\mathbf{E}$ due to a finite line charge at a point in space is given by $$ E = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{r} \sin\theta $$ where $r$ is the perpendicular distance from the line to the point, and $\theta$ is the half-angle subtended by the line at the point. Here, $\epsilon_0=8.85\times10^{-12}\ \text{C}^2/\text{N}\cdot\text{m}^2$. 3. **Geometry and angles:** - The point is midway between the lines horizontally, so horizontal distance from each line to the point is $x=\frac{d}{2}=0.03\ \text{m}$. - The vertical distance from the line to the point is $y=0.40\ \text{m}$. - The perpendicular distance from the point to each line is $r=\sqrt{x^2 + y^2} = \sqrt{0.03^2 + 0.40^2} = \sqrt{0.0009 + 0.16} = \sqrt{0.1609} \approx 0.4011\ \text{m}$. 4. **Calculate angle $\theta$:** - Half-length of the line is $L/2=0.03\ \text{m}$. - The angle $\theta$ satisfies $\tan\theta = \frac{L/2}{r} = \frac{0.03}{0.4011} \approx 0.0748$. - So, $\theta = \arctan(0.0748) \approx 0.0746\ \text{rad}$. 5. **Calculate electric field magnitude from one line:** $$ E_1 = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{r} \sin\theta $$ Substitute values: $$ \frac{1}{4\pi\epsilon_0} = 9.0 \times 10^9\ \text{N}\cdot\text{m}^2/\text{C}^2 $$ $$ E_1 = 9.0 \times 10^9 \times \frac{2 \times 1.00 \times 10^{-6}}{0.4011} \times \sin(0.0746) $$ Calculate $\sin(0.0746) \approx 0.0745$: $$ E_1 = 9.0 \times 10^9 \times \frac{2 \times 10^{-6}}{0.4011} \times 0.0745 = 9.0 \times 10^9 \times 4.987 \times 10^{-6} \times 0.0745 = 9.0 \times 10^9 \times 3.715 \times 10^{-7} = 3343.5\ \text{N/C} $$ 6. **Direction of fields:** - The left line is negatively charged, so its field points toward the line. - The right line is positively charged, so its field points away from the line. - At the midpoint, the horizontal components of the two fields add because they point in the same direction (from positive to negative line). - The vertical components cancel because they are equal and opposite. 7. **Calculate horizontal component of $E_1$:** - The angle between $\mathbf{E}_1$ and the horizontal is $\phi = \arctan(y/x) = \arctan(0.40/0.03) \approx 1.496\ \text{rad}$. - Horizontal component: $$ E_{1x} = E_1 \cos\phi = 3343.5 \times \cos(1.496) \approx 3343.5 \times 0.066 = 220.7\ \text{N/C} $$ 8. **Total electric field:** - Both lines contribute $E_{1x}$ in the same direction, so total horizontal field: $$ E_{total} = 2 \times 220.7 = 441.4\ \text{N/C} $$ - Vertical components cancel, so net field is horizontal, directed from positive to negative line. **Final answer:** $$ \boxed{E = 441\ \text{N/C} \quad \text{directed horizontally from the positive to the negative line}} $$