Question

The sum of two point charges is 6 microcoulomb. They attract each other with a force of 0.9N, when kept 40cm apart in vaccum. Calculate the charges.

Answer 1

From the Question, It is given that the charges attract each other. This means that the magnitude of both the charges differs in the sign because unlike changes attract each other. Thus, Let the magnitude of the two charges be q₁ and -q₂.

Given conditions ⇒

q₁ + (-q₂) = 6 μC

 = 6 × 10⁻⁶ C.

∴ q₁ = (6 × 10⁻⁶) + q₂

Force of the Repulsion between the charges(F) = 0.9 N.

Distance between the charges(r) = 40 cm.

= 0.4 m.

Now, Using the Coulomb's law,

F = k q₁q₂/r²

where, k (Coulomb law Constant)= 9 × 10⁹ N m² C⁻²

∴ 0.9 = [9 × 10⁹ × q₁q₂]/(0.4)²

∴ 0.9 × 0.16 = 9 × 10⁹ × q₁q₂

∴  q₁q₂ = 0.16 × 10⁻¹⁰

(6 × 10⁻⁶ + q₂)q₂ = 0.16 × 10⁻¹⁰

Let q₂ be a so that we will not face any inconvenience in the calculations.

∴ 6 × 10⁻⁶a + a² = 0.16 × 10⁻¹⁰

⇒ a² + 6 × 10⁻⁶a - 0.16 × 10⁻¹⁰ = 0

Now, this is the Quadratic Equation, so solving it by the Factorization method.

∴ a² + 6 × 10⁻⁶a - 16 × 10⁻¹² = 0 [This is done so that factors can be made easily.]

∴ a² - 2 × 10⁻⁶a + 8 × 10⁻⁶a - 16 × 10⁻¹² = 0

⇒ a (a - 2 × 10⁻⁶) + 8 × 10⁻⁶(a - 2 × 10⁻⁶) = 0

⇒ (a + 8 × 10⁻⁶)(a - 2 × 10⁻⁶) = 0

⇒ a = -8 × 10⁻⁶ and a = 2 × 10⁶

∴ a = -8 μC and a = 2 μC.

Now,

q₁ = 6 μC + q₂ (or a)

In First Case,

q₁ = 6 + (-8)

= -2 μC

Also,

q₁ = 6 + 2

= 8 μC.

Hope it helps.