Let's solve the problem step by step, with clear explanation blocks and a final answer summary. --- ### **Definition of Parameters** - **q₁:** Charge at \( x = -a \) - **q₂:** Charge at \( x = +a \) - **Q:** Charge at \( x = x_3 \) (where net force on Q must be zero) - **a:** Given distance from origin - **x₃:** Position of Q (will use values \( +0.474a \) and \( +1.53a \)) - **F:** Electrostatic force --- ### **Step 1: Write the Force Equations** #### **Explanation** The net electrostatic force on Q (at \( x_3 \)) is the sum of the forces due to \( q_1 \) and \( q_2 \). Each force is calculated using Coulomb's Law: \[ F = k \frac{qQ}{r^2} \] where \( r \) is the distance between charges. Set the net force on Q to zero: \[ F_{13} + F_{23} = 0 \] - \( F_{13} \) is the force on Q by \( q_1 \) (at \( x = -a \)) - \( F_{23} \) is the force on Q by \( q_2 \) (at \( x = +a \)) Calculate the distances: - Distance from Q to \( q_1 \): \( x_3 - (-a) = x_3 + a \) - Distance from Q to \( q_2 \): \( x_3 - a \) Write the equation: \[ k \frac{q_1 Q}{(x_3 + a)^2} + k \frac{q_2 Q}{(x_3 - a)^2} = 0 \] --- ### **Step 2: Solve for the Ratio \( q_1/q_2 \) at Each Position** #### **Explanation** \( k \) and \( Q \) cancel out (assuming nonzero). Rearranging gives: \[ \frac{q_1}{(x_3 + a)^2} = -\frac{q_2}{(x_3 - a)^2} \] \[ \frac{q_1}{q_2} = -\frac{(x_3 + a)^2}{(x_3 - a)^2} \] Substitute \( x_3 \) for each part: #### (a) For \( x_3 = +0.474a \): \[ \frac{q_1}{q_2} = -\frac{(0.474a + a)^2}{(0.474a - a)^2} = -\frac{(1.474a)^2}{(-0.526a)^2} = -\frac{(1.474)^2}{(0.526)^2} = -\frac{2.172}{0.277} = -7.84 \] #### (b) For \( x_3 = +1.53a \): \[ \frac{q_1}{q_2} = -\frac{(1.53a + a)^2}{(1.53a - a)^2} = -\frac{(2.53a)^2}{(0.53a)^2} = -\frac{6.401}{0.2809} = -22.77 \] --- ### **Final Answer Summary** - **(a) When \( x = +0.474a \):** \[ \boxed{-7.84} \] - **(b) When \( x = +1.53a \):** \[ \boxed{-22.77} \] **The negative sign indicates that \( q_1 \) and \( q_2 \) must have opposite signs.**

Let's work through this problem step-by-step, including clear explanations at each stage, to determine the ratio \( \frac{q_1}{q_2} \) when the net electrostatic force on particle 3 is zero at specified positions. --- ### **Step 1: Understand and set up the force balance equation** #### **Explanation:** - Particles \( q_1 \) and \( q_2 \) are fixed at positions \( x = -a \) and \( x = +a \), respectively. - Particle 3 with charge \( Q \) is located at position \( x_3 \). - The electrostatic force on \( Q \) due to \( q_1 \) and \( q_2 \) must sum to zero for the net force to be zero: \[ F_{13} + F_{23} = 0 \] - Coulomb's Law states: \[ F = k \frac{q \times Q}{r^2} \] - The forces are along the x-axis, so their magnitudes are proportional to the charges and inversely proportional to the square of the distances. #### **Distance calculations:** - Distance from \( q_1 \) at \( x = -a \) to \( Q \) at \( x = x_3 \): \[ r_1 = |x_3 - (-a)| = |x_3 + a| \] - Distance from \( q_2 \) at \( x = +a \) to \( Q \): \[ r_2 = |x_3 - a| \] --- ### **Step 2: Derive the ratio \( \frac{q_1}{q_2} \) for zero net force** #### **Explanation:** - The forces \( F_{13} \) and \( F_{23} \) act along the x-axis, and their directions depend on the signs of the charges. - For the net force to be zero, the magnitudes must satisfy: \[ \left| F_{13} \right| = \left| F_{23} \right| \] - Since the force direction depends on the signs, the relation between \( q_1 \) and \( q_2 \) involves their signs and magnitudes. - The force balance equation: \[ \frac{|q_1|}{r_1^2} = \frac{|q_2|}{r_2^2} \] - To find the ratio \( \frac{q_1}{q_2} \), include signs: \[ \frac{q_1}{q_2} = - \frac{r_1^2}{r_2^2} \] The negative sign accounts for opposite directions of forces when charges are of opposite signs. --- ### **Calculations for each position:** **(a) When \( x_3 = +0.474a \):** - \( r_1 = |0.474a + a| = 1.474a \) - \( r_2 = |0.474a - a| = 0.526a \) Calculate the ratio: \[ \frac{q_1}{q_2} = - \frac{(1.474a)^2}{(0.526a)^2} = - \frac{(1.474)^2}{(0.526)^2} \] \[ = - \frac{2.172}{0.277} \approx -7.84 \] --- **(b) When \( x_3 = +1.53a \):** - \( r_1 = |1.53a + a| = 2.53a \) - \( r_2 = |1.53a - a| = 0.53a \) Calculate the ratio: \[ \frac{q_1}{q_2} = - \frac{(2.53a)^2}{(0.53a)^2} = - \frac{6.401}{0.2809} \approx -22.77 \] --- ### **Final Summary:** | Position of particle 3 | \( \frac{q_1}{q_2} \) | |--------------------------|---------------------| | \( x_3 = +0.474a \) | **-7.84** | | \( x_3 = +1.53a \) | **-22.77** | The negative sign indicates that \( q_1 \) and \( q_2 \) must have opposite signs to produce zero net force at those positions. --- **Q:** When the net electrostatic force on particle 3 is zero, what must be the ratio \( \frac{q_1}{q_2} \) at the given positions? **A:** - At \( x = +0.474a \): \( \boxed{-7.84} \) - At \( x = +1.53a \): \( \boxed{-22.77} \)