Question

Two identical bodies are placed 4.5m apart on a table. Force of repulsion between them is 49/9 N. Find the value of charge on each hbody

Answer 1

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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