Question

Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m. The charge on each of the sphere is \(\frac{a}{21} \times 10^{-8} C\). 
The value of 'a' will be ______. [Given g = 10 ms–2]

Answer 1

Each sphere has mass m = 10 mg = 10 × 10⁻⁶ kg. Length of thread = 0.5 m, and separation between spheres = 0.20 m.

Step 1: Geometry
Each sphere is displaced by 0.1 m horizontally, so sinθ = 0.1 / 0.5 = 0.2 ⇒ cosθ = √(1 − 0.04) = √0.96 = 0.98 ⇒ tanθ = 0.2 / 0.98 = 0.1 / √24

Step 2: Vertical equilibrium
T cosθ = mg mg = 10 × 10⁻⁶ × 10 = 10⁻⁴ N …(i)

Step 3: Horizontal equilibrium
T sinθ = F_e Coulomb force: F_e = k q² / r² Here r = 0.20 m, thus r² = 0.04 So, T sinθ = (9 × 10⁹ q²) / 0.04 …(ii)

Step 4: Divide (ii) by (i)
(T sinθ) / (T cosθ) = tanθ ⇒ tanθ = (9 × 10⁹ q² / 0.04) / (mg)

Substitute values:
tanθ = 0.1 / √24 mg = 10⁻⁴ Therefore,
0.1 / √24 = (9 × 10⁹ q² / 0.04) / 10⁻⁴

Step 5: Solve for q
q² = (0.1 / √24) × 0.04 × 10⁻⁴ / (9 × 10⁹) q = √(2√10 / (3√24)) × 10⁻⁸ Simplifying gives:
q ≈ 0.95 × 10⁻⁸ C

Step 6: Compare with given form
q = (a / 21) × 10⁻⁸ ⇒ a / 21 = 0.95 ⇒ a ≈ 20

Final Answer: a = 20

Answer 2