Question

the electrostatic force of repulsion between two positively charged ions carrying equal charge is 7.4×10^-9N, when they are separated by a distance of 5×10^-10m . What is the value of charge ? How many electrons are missing from each ion?​

Answer 1

Answer

Number of missing electron from each ion is 3

Explanation :

GIVEN : The electrostatic forve of repulsion between two positively charged ions carrying equal charge is 7.4× 10⁻⁹ N

Distance of separation , d = 5× 10⁻¹⁰m = 5A⁰

We have to find the number of electron mission from each ions

Let us suppose charges is q₁ = q₂ = q (given equal charges)

Let n be the number of electron missing

Using columb's law F = kq²/r²

q²= F× r² /k

q²= 7.4 × 10⁻⁹ × (5× 10⁻¹⁰) ²/9× 10⁹

q²= 7.4×25× 10⁻²⁹/9×10⁹

q² = 20.56 × 10⁻³⁸ /9

q = √20.56 ×10³⁸ ≈ 4.5×10⁻¹⁹ C

as q = ne

so, n = q/e = 4.5×10⁻¹⁹/1.6×10⁻¹⁹ [∵Charge of electrone is 1.6×10⁻⁹C]

n = 2.81 roundly n = 3

Since, Number of electron missing from each ion is 3 .

Answer 2

Answer:

n=2.81 roundly n=3

since, number of electron missing from each ion is 3