# Problem Restatement - **Charge \( q_1 \):** \( +1.61~\mu C \) at \( x_1 = .~m \) - **Charge \( q_2 \):** \( -3.57~\mu C \) at \( x_2 = 12.7~m \) - **Test charge \( q_ \):** \( 3.32~\mu C \) placed somewhere on the x-axis - **Question:** Where on the x-axis (other than infinity) is the **net electric force on \( q_ \) zero**? --- # Step 1: Reasoning About Regions The force on \( q_ \) from each charge is: \[ F_1 = k \frac{|q_1 q_|}{r_1^2} \] \[ F_2 = k \frac{|q_2 q_|}{r_2^2} \] where \( r_1 \) and \( r_2 \) are the distances from \( q_1 \) and \( q_2 \) respectively. Since \( q_1 \) is **positive** and \( q_2 \) is **negative**, their forces on a positive test charge are in **opposite directions**. - **Between the charges (\( < x < 12.7 \))**: Both forces are in the **same direction** (towards \( q_2 \)), so they cannot cancel. - **Left of both (\( x < \))**: Forces can oppose and cancel. - **Right of both (\( x > 12.7 \))**: Forces can oppose and cancel. --- # Step 2: Write the Force Balance Equation Let \( x \) be the location of \( q_ \): - Distance to \( q_1 \): \( |x| \) - Distance to \( q_2 \): \( |x - 12.7| \) ### Region 1: \( x < \) - Force from \( q_1 \): repulsive, rightward. - Force from \( q_2 \): attractive, leftward. Set magnitudes equal for equilibrium: \[ F_1 = F_2 \] \[ k \frac{q_1 q_}{(x)^2} = k \frac{|q_2| q_}{(x - 12.7)^2} \] \( k \) and \( q_ \) cancel (since \( q_ > \)), so: \[ \frac{q_1}{x^2} = \frac{|q_2|}{(x - 12.7)^2} \] \[ \frac{1.61}{x^2} = \frac{3.57}{(x - 12.7)^2} \] --- # Step 3: Solve for \( x \) Cross-multiplied: \[ 1.61 (x - 12.7)^2 = 3.57 x^2 \] Take square roots: \[ \frac{x - 12.7}{x} = \pm \sqrt{\frac{3.57}{1.61}} \] But let's expand and solve as a quadratic: \[ 1.61 (x^2 - 25.4x + 161.29) = 3.57 x^2 \] \[ 1.61x^2 - 40.894x + 259.677 = 3.57x^2 \] \[ = 3.57x^2 - 1.61x^2 + 40.894x - 259.677 \] \[ = 1.96x^2 + 40.894x - 259.677 \] --- # Step 4: Quadratic Formula \[ 1.96x^2 + 40.894x - 259.677 = \] Use \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1.96 \), \( b = 40.894 \), \( c = -259.677 \). Calculate discriminant: \[ D = (40.894)^2 - 4 \times 1.96 \times (-259.677) \] \[ = 1672.468 + 2036.64 = 3709.108 \] \[ x = \frac{-40.894 \pm \sqrt{3709.108}}{3.92} \] \[ \sqrt{3709.108} \approx 60.892 \] So, \[ x_1 = \frac{-40.894 + 60.892}{3.92} \approx \frac{19.998}{3.92} \approx 5.10 \] \[ x_2 = \frac{-40.894 - 60.892}{3.92} \approx \frac{-101.786}{3.92} \approx -25.96 \] --- # Step 5: Interpret Solutions - \( x_1 = 5.10~\text{m} \): **Between the charges**. In this region, both forces are in the same direction (see above), so this solution is **not physical**. - \( x_2 = -25.96~\text{m} \): **Left of both charges**. This is the valid position. --- # Step 6: Check the \( x > 12.7 \) Region If you try the same calculation for \( x > 12.7 \), the force directions are the same (both toward the negative charge), so no equilibrium is possible. --- # Final Answer ## The location on the x-axis where the electric force on the third charge is zero: \[ \boxed{x = -25.96~\text{m}} \] **to the left of both charges.** --- **Summary Table:** | Variable | Value | |-------------|------------------| | \( q_1 \) | \( +1.61~\mu C \) at \( x = .~m \) | | \( q_2 \) | \( -3.57~\mu C \) at \( x = 12.7~m \) | | \( q_ \) | \( 3.32~\mu C \) | | Zero-force location | \( x = -25.96~m \) |