Let's break down the problem step by step. ## **1. Charges and their Placement** Let’s assume the charges are arranged at the corners of a square of side \( a \), labeled as follows: - Lower left: \( q \) (this is the charge we're interested in, call this point \( Q \)) - Lower right: \( B = 5.5q \) - Upper right: \( C = 4.5q \) - Upper left: \( q_4 \) (not specified, so assume it's not relevant or is zero) We’ll assume the origin \((0,0)\) is at the lower left corner where \(q\) is placed. So: - Charge at \( Q(0,0) \): \( q \) - Charge at \( B(a,0) \): \( 5.5q \) - Charge at \( C(a,a) \): \( 4.5q \) - Charge at \( D(0,a) \) (let’s call this \( q_4 \), not specified) Assume \( q_4 = 0 \), unless told otherwise. --- ## **2. Calculate Forces on \( q \) at \( (0,0) \)** The electric force on \( q \) due to another charge \( q' \) at position \( (x',y') \) is: \[ \vec{F} = k_e \frac{qq'}{r^2} \hat{r} \] where \( \vec{r} \) is the displacement vector from \( q \) to \( q' \). Let’s calculate the force due to each charge. --- ### **A. Due to \( B = 5.5q \) at \( (a,0) \):** - Displacement: \( \vec{r}_B = (a,0) \) - Distance: \( r_B = a \) - Direction: To the right (\( \hat{i} \)) \[ \vec{F}_B = k_e \frac{q \cdot 5.5q}{a^2} \hat{i} = 5.5 k_e \frac{q^2}{a^2} \hat{i} \] --- ### **B. Due to \( C = 4.5q \) at \( (a,a) \):** - Displacement: \( \vec{r}_C = (a,a) \) - Distance: \( r_C = \sqrt{a^2 + a^2} = a\sqrt{2} \) - Direction: \( (\hat{i} + \hat{j})/\sqrt{2} \) \[ \vec{F}_C = k_e \frac{q \cdot 4.5q}{(a\sqrt{2})^2} \frac{\vec{r}_C}{r_C} = k_e \frac{4.5q^2}{2a^2} \cdot \frac{1}{\sqrt{2}} (\hat{i} + \hat{j}) = \frac{4.5}{2\sqrt{2}} k_e \frac{q^2}{a^2} (\hat{i} + \hat{j}) = \frac{2.25}{\sqrt{2}} k_e \frac{q^2}{a^2} (\hat{i} + \hat{j}) \] --- ### **C. Due to \( D \) at \( (0,a) \):** If \( q_4 \) is present, call it \( q_D \). If not specified, ignore. --- ## **3. Add the Forces** Sum the \( \hat{i} \) and \( \hat{j} \) components: ### **i-component:** \[ F_x = 5.5 k_e \frac{q^2}{a^2} + \frac{2.25}{\sqrt{2}} k_e \frac{q^2}{a^2} \] ### **j-component:** \[ F_y = \frac{2.25}{\sqrt{2}} k_e \frac{q^2}{a^2} \] --- ## **4. Resultant Force (Magnitude and Direction)** ### **Magnitude:** \[ |\vec{F}| = \sqrt{F_x^2 + F_y^2} \] Plug in the values: Let’s write \( K = k_e \frac{q^2}{a^2} \) for simplicity. \[ F_x = 5.5K + \frac{2.25}{\sqrt{2}}K \] \[ F_y = \frac{2.25}{\sqrt{2}}K \] So: \[ |\vec{F}| = K \sqrt{\left(5.5 + \frac{2.25}{\sqrt{2}}\right)^2 + \left(\frac{2.25}{\sqrt{2}}\right)^2} \] Let’s simplify: Let \( \alpha = \frac{2.25}{\sqrt{2}} \) So, \[ |\vec{F}| = K \sqrt{(5.5 + \alpha)^2 + \alpha^2} \] --- ### **Direction (Angle from +x-axis):** \[ \tan\theta = \frac{F_y}{F_x} = \frac{\alpha}{5.5 + \alpha} \] \[ \theta = \arctan\left( \frac{\alpha}{5.5 + \alpha} \right ) \] --- ### **Final Answers in Terms of \( k_e, q, a \):** \[ \boxed{ \begin{aligned} |\vec{F}| &= k_e \frac{q^2}{a^2} \sqrt{ \left[5.5 + \frac{2.25}{\sqrt{2}}\right]^2 + \left[\frac{2.25}{\sqrt{2}}\right]^2 } \\ \theta &= \arctan\left( \frac{2.25/\sqrt{2}}{5.5 + 2.25/\sqrt{2}} \right) \end{aligned} } \] Where \(\theta\) is measured **above the +x-axis** (to the right). --- #### **If you want decimal approximations:** - \( 2.25/\sqrt{2} \approx 1.591 \) - \( 5.5 + 1.591 = 7.091 \) - So, \[ |\vec{F}| \approx k_e \frac{q^2}{a^2} \sqrt{(7.091)^2 + (1.591)^2} = k_e \frac{q^2}{a^2} \sqrt{50.315 + 2.532} = k_e \frac{q^2}{a^2} \sqrt{52.847} \approx 7.27\, k_e \frac{q^2}{a^2} \] \[ \theta \approx \arctan\left( \frac{1.591}{7.091} \right) \approx \arctan(0.224) \approx 12.6^\circ \] --- ## **SUMMARY:** \[ \boxed{ |\vec{F}| = k_e \frac{q^2}{a^2} \sqrt{ \left[5.5 + \frac{2.25}{\sqrt{2}}\right]^2 + \left[\frac{2.25}{\sqrt{2}}\right]^2 } } \] \[ \boxed{ \theta = \arctan\left( \frac{2.25/\sqrt{2}}{5.5 + 2.25/\sqrt{2}} \right) } \] Where \( \theta \) is **above the +x-axis (to the right)**. Let me know if you need the symbolic answer or the decimal approximation!