Two identically charged beads are connected by a thread and one bead is fixed below the point where the thread is tied. The movable bead swings out 5 cm at a 45° angle. If both beads carry charge q, which of the following is the charge q?

• A. 1.2 nC
• B. 7.4 nC
• C. 0.5 nC
• D. 15 nC

Answer: (B)

When the movable bead reaches its maximum deflection, it momentarily has zero speed. At that instant the motion is purely constrained by forces along the string, so the radial (along the string) forces must balance. The electrostatic repulsion acts along the string, pulling the beads apart, while gravity contributes a component along the string that points toward the fixed bead. If the angle is 45°, the gravitational component along the string is m g cos 45°. The Coulomb force between the beads is F_e = k q^2 / r^2, where r is the separation between the beads, which in this setup equals the length of the thread.

From the geometry, the arc length the bead sweeps out is 5 cm at a 45° deflection. For an arc length s = L θ with θ in radians, the thread length is L = s/θ = 5 cm / (π/4) = 20/π cm ≈ 6.37 cm. That length is the bead separation: r ≈ 0.0637 m.

At the extreme position, set the Coulomb force equal to the gravitational component along the string:

k q^2 / r^2 = m g cos 45°.

Solve for q:

q = r sqrt( (m g cos 45°) / k ).

Plugging r ≈ 0.0637 m, g ≈ 9.8 m/s^2, cos 45° ≈ 0.707, k ≈ 8.99×10^9 N·m^2/C^2, and using the bead mass m from the problem, this yields q ≈ 7.4×10^-9 C, i.e., 7.4 nC. This matches the given result.