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Q. Two Identical coins having similar charges are placed 4.5 m apart on a table. Force of repulsion between them is 49/9 N. The value of charge on each coin is :-1) 100 mC2) 200 mC3) 300 mC4) 35√10 C |
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Answer» LET charge on each COIN is q GIVEN, Force = 49/9 N Speration between charges , r = 4.5m ∵ Force = KQq/r² [ from Coulombs law] ∴ 49/9 = 9 × 10⁹ × q × q/(4.5)² ⇒ 49/9 = 9 × 10⁹ × q²/(4.5)² ⇒ 49 × (4.5)² × 10⁻⁹/9 × 9 = q² ⇒(7)² × (4.5)² × 10 × 10⁻¹⁰/(9)² = q² Taking square root both sides, ⇒ 7 × 4.5 × √10/9 × 10⁻⁵ = q ⇒ 3.5√10 × 10⁻⁵ = q ⇒ 35√10 × 10⁻⁶ = q Hence, charge on each coin is 35√10 μC It seems you did mistake in typing. Answer should be 35√10 μC , CORRECT Option is (4) 35√10 μC |
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