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]
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]
Approach Solution - 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 = Fe Coulomb force: Fe = 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
Approach Solution -2
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