Question:
Two identical balls each of mass $\frac{40}{3}$ g carry equal and opposite charge. They are suspended from a horizontal plate by silk threads each of length 1 m with a separation of 1.5 m between points of suspension. At equilibrium, if the distance between the balls is 30 cm then magnitude of charge on each ball is (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)
Two identical balls each of mass $\frac{40}{3}$ g carry equal and opposite charge. They are suspended from a horizontal plate by silk threads each of length 1 m with a separation of 1.5 m between points of suspension. At equilibrium, if the distance between the balls is 30 cm then magnitude of charge on each ball is (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)
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Geometry is key: find the horizontal offset per string $x = \frac{1.5 - 0.3}{2} = 0.6\text{ m}$. With $L=1\text{m}$, this forms a classic 3-4-5 right triangle ($0.6, 0.8, 1$), so $\tan\theta = 3/4$.
Updated On: Jun 3, 2026
- $10^{-5} \text{ C}$
- $10^{-6} \text{ C}$
- $10^{-3} \text{ C}$
- $2\times 10^{-6} \text{ C}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
For a suspended charged particle in static equilibrium, the balance of horizontal electrostatic force ($F_e$) and vertical gravitational force ($mg$) leads to the condition: $\tan\theta = \frac{F_e}{mg}$, where $F_e = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2}$.
Step 2: Meaning
Let the distance between the points of suspension be $D = 1.5 \text{ m}$ and the distance between the balls at equilibrium be $r = 30 \text{ cm} = 0.3 \text{ m}$. The horizontal displacement of each thread from its vertical position is $x = \frac{D - r}{2} = \frac{1.5 - 0.3}{2} = 0.6 \text{ m}$.
Step 3: Analysis
Since the string length is $L = 1 \text{ m}$, the vertical height is $h = \sqrt{L^2 - x^2} = \sqrt{1^2 - (0.6)^2} = 0.8 \text{ m}$. Thus, $\tan\theta = \frac{x}{h} = \frac{0.6}{0.8} = \frac{3}{4}$. Given $m = \frac{40}{3}\text{ g} = \frac{40}{3} \times 10^{-3} \text{ kg}$ and $g = 10 \text{ ms}^{-2}$, we can equate the forces: $\frac{F_e}{mg} = \tan\theta \implies F_e = mg \tan\theta = \left(\frac{40}{3} \times 10^{-3}\right) \times 10 \times \frac{3}{4} = 0.1 \text{ N}$. Now set up Coulomb's equation: $9 \times 10^9 \times \frac{q^2}{(0.3)^2} = 0.1 \implies 9 \times 10^9 \times \frac{q^2}{0.09} = 0.1 \implies 10^{11} \cdot q^2 = 0.1 \implies q^2 = 10^{-12} \implies q = 10^{-6} \text{ C}$.
Step 4: Conclusion
Therefore, the absolute magnitude of the electric charge on each ball is $10^{-6} \text{ C}$.
Final Answer: (B)
For a suspended charged particle in static equilibrium, the balance of horizontal electrostatic force ($F_e$) and vertical gravitational force ($mg$) leads to the condition: $\tan\theta = \frac{F_e}{mg}$, where $F_e = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2}$.
Step 2: Meaning
Let the distance between the points of suspension be $D = 1.5 \text{ m}$ and the distance between the balls at equilibrium be $r = 30 \text{ cm} = 0.3 \text{ m}$. The horizontal displacement of each thread from its vertical position is $x = \frac{D - r}{2} = \frac{1.5 - 0.3}{2} = 0.6 \text{ m}$.
Step 3: Analysis
Since the string length is $L = 1 \text{ m}$, the vertical height is $h = \sqrt{L^2 - x^2} = \sqrt{1^2 - (0.6)^2} = 0.8 \text{ m}$. Thus, $\tan\theta = \frac{x}{h} = \frac{0.6}{0.8} = \frac{3}{4}$. Given $m = \frac{40}{3}\text{ g} = \frac{40}{3} \times 10^{-3} \text{ kg}$ and $g = 10 \text{ ms}^{-2}$, we can equate the forces: $\frac{F_e}{mg} = \tan\theta \implies F_e = mg \tan\theta = \left(\frac{40}{3} \times 10^{-3}\right) \times 10 \times \frac{3}{4} = 0.1 \text{ N}$. Now set up Coulomb's equation: $9 \times 10^9 \times \frac{q^2}{(0.3)^2} = 0.1 \implies 9 \times 10^9 \times \frac{q^2}{0.09} = 0.1 \implies 10^{11} \cdot q^2 = 0.1 \implies q^2 = 10^{-12} \implies q = 10^{-6} \text{ C}$.
Step 4: Conclusion
Therefore, the absolute magnitude of the electric charge on each ball is $10^{-6} \text{ C}$.
Final Answer: (B)
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