Let's solve both parts step-by-step: --- ### **Given:** - \( Q_1 = 6.87 \) nC at \( x = 0 \) - \( Q_2 = 2.77 \) nC at \( x = 2.55 \) m - \( q = -7.98 \) nC placed at position \( x \) - Find \( x \) such that the net force on \( q \) is zero. --- ## **Step 1: Write the Forces** The force on \( q \) by \( Q_1 \): \[ F_1 = k \frac{|Q_1 q|}{x^2} \] Direction: If \( x > 0 \), \( F_1 \) will be attractive (since \( Q_1 > 0 \), \( q < 0 \)), so toward the origin. The force on \( q \) by \( Q_2 \): \[ F_2 = k \frac{|Q_2 q|}{(x - 2.55)^2} \] Direction: If \( x < 2.55 \), \( F_2 \) will be attractive (since both \( Q_2 > 0 \) and \( q < 0 \)), so toward \( x = 2.55 \). --- ## **Step 2: Set the Net Force to Zero** \[ F_1 + F_2 = 0 \] But remember to consider directions carefully. Let’s place \( q \) at position \( x \). ### **Case 1: \( x < 0 \) (to the left of both charges)** - Both forces point to the right (positive direction). ### **Case 2: \( 0 < x < 2.55 \) (between the charges)** - \( F_1 \): toward origin (left, negative direction) - \( F_2 \): toward \( x = 2.55 \) (right, positive direction) ### **Case 3: \( x > 2.55 \) (right of both charges)** - Both forces point to the left (negative direction). So only in **Case 2** can the forces balance each other. --- ## **Step 3: Set Up the Equation for \( 0 < x < 2.55 \)** The magnitude of the forces must be equal and opposite: \[ k \frac{|Q_1 q|}{x^2} = k \frac{|Q_2 q|}{(x - 2.55)^2} \] \( k \) and \( |q| \) cancel out: \[ \frac{Q_1}{x^2} = \frac{Q_2}{(x - 2.55)^2} \] \[ Q_1 (x - 2.55)^2 = Q_2 x^2 \] \[ Q_1 (x^2 - 2 \cdot 2.55 x + 2.55^2) = Q_2 x^2 \] \[ Q_1 x^2 - 2 Q_1 \cdot 2.55 x + Q_1 \cdot 2.55^2 = Q_2 x^2 \] \[ Q_1 x^2 - Q_2 x^2 = 2 Q_1 \cdot 2.55 x - Q_1 \cdot 2.55^2 \] \[ (Q_1 - Q_2)x^2 - 2 Q_1 \cdot 2.55 x + Q_1 \cdot 2.55^2 = 0 \] --- ## **Step 4: Plug in the Values** - \( Q_1 = 6.87 \) - \( Q_2 = 2.77 \) - \( 2.55^2 = 6.5025 \) So, \[ (6.87 - 2.77)x^2 - 2 \times 6.87 \times 2.55 x + 6.87 \times 6.5025 = 0 \] \[ 4.10 x^2 - 35.037 x + 44.701 = 0 \] --- ## **Step 5: Solve the Quadratic Equation** \[ 4.10 x^2 - 35.037 x + 44.701 = 0 \] Use quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 4.10 \), \( b = -35.037 \), \( c = 44.701 \): \[ x = \frac{35.037 \pm \sqrt{(-35.037)^2 - 4 \cdot 4.10 \cdot 44.701}}{2 \cdot 4.10} \] Calculate the discriminant: \[ (-35.037)^2 = 1227.59 \] \[ 4 \cdot 4.10 \cdot 44.701 = 733.82 \] \[ \text{Discriminant} = 1227.59 - 733.82 = 493.77 \] \[ \sqrt{493.77} \approx 22.23 \] So, \[ x_1 = \frac{35.037 + 22.23}{8.2} = \frac{57.267}{8.2} \approx 6.98 \, \text{m} \] \[ x_2 = \frac{35.037 - 22.23}{8.2} = \frac{12.807}{8.2} \approx 1.56 \, \text{m} \] **Recall \( 0 < x < 2.55 \) is the valid region.** **Final Answer (Part 1):** \[ \boxed{1.56 \, \text{m}} \] --- ## **Part 2: If \( Q_2 \) is Negative** Now \( Q_2 = -2.77 \) nC. Set up the equation as before, but now the force from \( Q_2 \) reverses direction (repulsive since both are negative). The only possible region for equilibrium is now **outside the segment**, specifically to the **right of both charges** (\( x > 2.55 \)), since both charges are negative and will both repel \( q \) (which is negative). So for \( x > 2.55 \): \[ k \frac{|Q_1 q|}{x^2} = k \frac{|Q_2 q|}{(x - 2.55)^2} \] but now \( Q_2 \) is negative, so direction is away from both \( Q_1 \) and \( Q_2 \). Set up the same quadratic as before, but with \( Q_2 = -2.77 \): \[ (Q_1 - Q_2)x^2 - 2 Q_1 \cdot 2.55 x + Q_1 \cdot 2.55^2 = 0 \] \[ (6.87 - (-2.77))x^2 - 2 \times 6.87 \times 2.55 x + 6.87 \times 6.5025 = 0 \] \[ 9.64 x^2 - 35.037 x + 44.701 = 0 \] Now solve: \[ x = \frac{35.037 \pm \sqrt{(-35.037)^2 - 4 \cdot 9.64 \cdot 44.701}}{2 \cdot 9.64} \] \[ 4 \cdot 9.64 \cdot 44.701 = 1721.25 \] \[ (-35.037)^2 = 1227.59 \] \[ \text{Discriminant} = 1227.59 - 1721.25 = -493.66 \] Negative discriminant—no real solution for \( x > 2.55 \) (to the right). Let's check to the **left** of both charges (\( x < 0 \)), try the other region: But in this scenario, both will repel, so force cannot be balanced for \( q \) anywhere. **Therefore, for \( Q_2 \) negative, there is NO POSITION where the net force on \( q \) is zero.** But let's double-check the setup: Actually, for \( x < 0 \): - \( Q_1 \) repels \( q \) to the left. - \( Q_2 \) repels \( q \) to the left (since both are negative and \( q \) is negative). So the only possible region is between \( Q_1 \) and \( Q_2 \) (\( 0 < x < 2.55 \)), but here, both forces are in the same direction (both away from their respective charges), so there's no balance. But let's write the force directions explicitly in the region \( x > 2.55 \): - Both forces are *toward* the charges (since \( q \) is negative and both \( Q_1, Q_2 \) are negative): both forces point left (negative direction). No equilibrium point exists with both charges negative and \( q \) negative. --- ## **Final Answers:** 1. **With \( Q_2 = +2.77 \) nC:** \( \boxed{1.56 \, \text{m}} \) from the origin (between the two charges). 2. **With \( Q_2 = -2.77 \) nC:** \( \boxed{\text{No position exists where the net force is zero}} \). --- **Let me know if you want a more detailed discussion or a diagram!**