Let's solve the problem **step by step**. --- ## **Given Data** - \( q_1 = 100\,\text{nC} = 100 \times 10^{-9}\,\text{C} \) - \( q_2 = -200\,\text{nC} = -200 \times 10^{-9}\,\text{C} \) - \( q_3 = +300\,\text{nC} = 300 \times 10^{-9}\,\text{C} \) - \( b = 5.\,\text{cm} = .050\,\text{m} \) - \( c = 7.\,\text{cm} = .070\,\text{m} \) - **Coulomb's Constant:** \( k = \frac{1}{4\pi\epsilon_} = 8.99 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2 \) We need to find the **net electric force on \( q_2 \)**. --- ## **Step 1: Find Forces Acting on \( q_2 \)** ### **A. Force on \( q_2 \) due to \( q_1 \) (\( \vec{F}_{21} \))** - Distance between \( q_2 \) and \( q_1 \) is \( b \). - Direction: \( q_1 \) is above \( q_2 \) (along \( +y \)). \[ F_{21} = k \frac{|q_2 q_1|}{b^2} \] Plug in the values: \[ F_{21} = 8.99 \times 10^9 \frac{|(-200 \times 10^{-9})(100 \times 10^{-9})|}{(.050)^2} \] \[ = 8.99 \times 10^9 \cdot \frac{200 \times 100 \times 10^{-18}}{.0025} \] \[ = 8.99 \times 10^9 \cdot \frac{20000 \times 10^{-18}}{.0025} \] \[ = 8.99 \times 10^9 \cdot \frac{2 \times 10^{-14}}{.0025} \] \[ = 8.99 \times 10^9 \cdot 8 \times 10^{-12} \] \[ = 71.92 \times 10^{-3} = .0719\,\text{N} \] - Since \( q_2 \) is negative and \( q_1 \) is positive, the force is **attractive** (points **upwards**, \( +y \)). --- ### **B. Force on \( q_2 \) due to \( q_3 \) (\( \vec{F}_{23} \))** - Distance between \( q_2 \) and \( q_3 \) is \( c \). - Direction: \( q_3 \) is to the right of \( q_2 \) (along \( +x \)). \[ F_{23} = k \frac{|q_2 q_3|}{c^2} \] \[ = 8.99 \times 10^9 \frac{|(-200 \times 10^{-9})(300 \times 10^{-9})|}{(.070)^2} \] \[ = 8.99 \times 10^9 \cdot \frac{200 \times 300 \times 10^{-18}}{.0049} \] \[ = 8.99 \times 10^9 \cdot \frac{60000 \times 10^{-18}}{.0049} \] \[ = 8.99 \times 10^9 \cdot \frac{6 \times 10^{-14}}{.0049} \] \[ = 8.99 \times 10^9 \cdot 1.2245 \times 10^{-11} \] \[ \approx .110\,\text{N} \] - The force is **attractive** (since \( q_2 \) is negative and \( q_3 \) is positive), so it points **right** (\( +x \)). --- ## **Step 2: Net Force (Vector Addition)** Both forces are **perpendicular** (\( F_{21} \) along \( +y \), \( F_{23} \) along \( +x \)). \[ F_{\text{net}} = \sqrt{F_{21}^2 + F_{23}^2} \] \[ = \sqrt{(.0719)^2 + (.110)^2} \] \[ = \sqrt{.00517 + .0121} \] \[ = \sqrt{.01727} \] \[ = .131\,\text{N} \] --- ### **Direction (Angle with Respect to +x Axis)** \[ \tan\theta = \frac{F_{21}}{F_{23}} = \frac{.0719}{.110} = .654 \] \[ \theta = \arctan(.654) \approx 33^\circ \] --- ## **Final Answer** \[ \boxed{ \text{The net electric force on } q_2 \text{ is } .13\,\text{N} \text{ at } 33^\circ \text{ above the } +x \text{ axis.} } \] --- ### **Summary Table** | Force | Magnitude (N) | Direction | |------------|---------------|-------------------| | \( F_{21} \) | .0719 | \( +y \) | | \( F_{23} \) | .110 | \( +x \) | | \( F_{\text{net}} \) | .13 | \( 33^\circ \) above \( +x \) | --- **Let me know if you need a diagram or further clarifications!**