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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[Explain properly with a solution]Plzz answer it aand fast

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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