Question:medium

Four electric charges 2 µC, Q, 4 µC and 12 µC are placed on x-axis at distances x = 0, 1 cm, 2 cm and 4 cm respectively. If the net force acting on the charge at origin is zero, then Q =

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When solving for an unknown charge in an equilibrium problem, first determine the required sign of the charge by analyzing the directions of the known forces. This prevents sign errors in the final answer. Also, in ratios like this, you can often keep non-SI units (like cm) as long as they are used consistently, as they will cancel out.
Updated On: Mar 26, 2026
  • -3.5 µC
  • -1.75 µC
  • -2.75 µC
  • -5.5 µC
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The Correct Option is B

Solution and Explanation

Step 1: Analyze Forces on Charge at Origin: Let the charge at origin be \( q_0 = 2 \, \mu\text{C} \). The forces acting on \( q_0 \) are due to \( Q \) at \( x_1=1 \), \( q_2=4 \, \mu\text{C} \) at \( x_2=2 \), and \( q_3=12 \, \mu\text{C} \) at \( x_3=4 \). According to Coulomb's Law, force \( F = \frac{k q_1 q_2}{r^2} \). Since \( q_0 \) is positive, repulsive forces (from positive charges) will be in the negative x-direction. For net force to be zero, attractive forces must balance repulsive ones.
Step 2: Equation for Net Force: Sum of forces \( = 0 \). \( \frac{k q_0 Q}{(1)^2} + \frac{k q_0 (4)}{(2)^2} + \frac{k q_0 (12)}{(4)^2} = 0 \) (Distances in cm, but units cancel out if consistent). Cancel \( k q_0 \) from all terms: \[ \frac{Q}{1} + \frac{4}{4} + \frac{12}{16} = 0 \]
Step 3: Solve for Q: \[ Q + 1 + \frac{3}{4} = 0 \] \[ Q + 1 + 0.75 = 0 \] \[ Q + 1.75 = 0 \] \[ Q = -1.75 \, \mu\text{C} \]
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