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Two identical small spheres carry charges $+10\,\mu\text{C}$ each and are placed $30\,\text{cm}$ apart in vacuum. The magnitude of the electrostatic force between them is closest to (Take $k = 9\times 10^9\,\text{N\,m}^2/\text{C}^2$.)
A$1\,\text{N}$
B$10\,\text{N}$
C$100\,\text{N}$
D$1000\,\text{N}$
Answer & Solution
Correct answer: B. $10\,\text{N}$
1. Coulomb's law: $F = k\,\dfrac{q_1 q_2}{r^2}$.
2. Substitute: $k = 9\times 10^9$; $q_1 = q_2 = 10\,\mu\text{C} = 10^{-5}\,\text{C}$; $r = 0.30\,\text{m}$.
3. Numerator: $q_1 q_2 = (10^{-5})^2 = 10^{-10}\,\text{C}^2$.
4. $k\,q_1 q_2 = (9\times 10^9)(10^{-10}) = 0.9\,\text{N\,m}^2$.
5. Denominator: $r^2 = (0.30)^2 = 0.09\,\text{m}^2$.
6. $F = 0.9/0.09 = 10\,\text{N}$.
7. Other options come from common unit slips: A forgets to square 10 µC; C drops a factor of 10; D treats µC as C.
_Source: NCERT Class 12 Physics Part 1, Ch 1, §1.5 (Coulomb's Law) + Example 1.1, p. 5–7._
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