Question

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 =

• A. -3.5 µC
• B. -1.75 µC
• C. -2.75 µC
• D. -5.5 µC

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves Coulomb's Law and the principle of superposition. The net force on a charge is the vector sum of the individual forces exerted on it by all other charges. For the net force to be zero, the vector sum of all forces must be zero.
Step 2: Key Formula or Approach:
Coulomb's Law: The force between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) is \(F = k \frac{q_1 q_2}{r^2}\).
Let the charge at the origin be \(q_0 = 2\) µC. The other charges are \(q_1 = Q\) at \(r_1 = 1\) cm, \(q_2 = 4\) µC at \(r_2 = 2\) cm, and \(q_3 = 12\) µC at \(r_3 = 4\) cm.
The net force on \(q_0\) is \(\vec{F}_{net} = \vec{F}_{01} + \vec{F}_{02} + \vec{F}_{03} = 0\).
Step 3: Detailed Explanation:
Let's analyze the direction of the forces on the charge \(q_0 = 2\) µC at the origin. Let the positive x-direction be to the right.
- Force due to \(q_2 = 4\) µC: Since \(q_0\) and \(q_2\) are both positive, the force \(\vec{F}_{02}\) is repulsive, pushing \(q_0\) to the left (in the -x direction).
- Force due to \(q_3 = 12\) µC: Since \(q_0\) and \(q_3\) are both positive, the force \(\vec{F}_{03}\) is also repulsive, pushing \(q_0\) to the left (in the -x direction).
- For the net force to be zero, the force due to \(q_1 = Q\), which is \(\vec{F}_{01}\), must balance the other two forces. This means \(\vec{F}_{01}\) must point to the right (in the +x direction).
- An attractive force is needed between \(q_0\) and \(Q\). Since \(q_0\) is positive, \(Q\) must be negative.
Now, we can equate the magnitudes of the forces:
\[ F_{01} = F_{02} + F_{03} \] \[ k \frac{|q_0 Q|}{r_1^2} = k \frac{q_0 q_2}{r_2^2} + k \frac{q_0 q_3}{r_3^2} \] We can cancel \(k\) and \(q_0\) from both sides. We can also use distances in cm, as the units will cancel out in the ratio.
\[ \frac{|Q|}{(1 \text{ cm})^2} = \frac{4 \text{ µC}}{(2 \text{ cm})^2} + \frac{12 \text{ µC}}{(4 \text{ cm})^2} \] \[ |Q| = \frac{4}{4} + \frac{12}{16} \] \[ |Q| = 1 + \frac{3}{4} = 1 + 0.75 = 1.75 \text{ µC} \] Since we determined that Q must be negative for the forces to balance, we have:
\[ Q = -1.75 \text{ µC} \] Step 4: Final Answer:
The value of charge Q is -1.75 µC. Therefore, option (B) is correct.