Question

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Answer 1

let the charges be q₁ and q₂ and given that q₁ + q₂ = 7μC

we know that according to coulombs law the force of attraction or repulsion between the two stationary charges q₁ and q₂ is given by F = 1/4πε₀ ( q₁×q₂)/r²

where ε₀ is the permittivity of free space and r is the distance between the two charges.

given that the force between the two charged particles is 1N and the distance is 0.3m and as the charges are in the free space, the value of 1/4πε₀ is 9 × 10⁹

substituting the above values in the coulombs law we get

1N = 9 × 10⁹× (q₁×q₂)/(0.3)²

solving we get (q₁×q₂) = 10⁻¹¹ C²

⇒ we know that 1μC = 10⁻⁶C

⇒ (q₁×q₂) can be written as (q₁×q₂) = 10 × 10⁻⁶C × 10⁻⁶C

⇒ (q₁×q₂) = 10 (μC)²

⇒ q₁ = 10 (μC)²/q₂

substituting the above q₁ = 10 (μC)²/q₂ in q₁ + q₂ = 7μC

we get 10 (μC)²/q₂ + q₂ = 7μC

let us make this simple by taking x = q₂

we get a quadratic equation x² - 7x + 10 = 0

solving the above quadratic equation we get x = 5 or x = 2

but we know that x = q₂

⇒ q₂ = 5μC or q₂ = 2μC

substituting the above q₂ values in q₁ + q₂ = 7μC

we get q₁ = 5μC or q₁ = 2μC

⇒ if q₁ = 5μC then q₂ = 2μC and if q₁ = 2μC then q₂ = 5μC