Question

Calculate coulomb force between two alpha particles separated by 3.2×10^-15m?

Answer 1

Charge of alpha particle = 2e
Here e = charge of electron

F = kq₁q₂/r²
= 4ke²/r²
= 4 × (9 × 10⁹) × (1.6 × 10⁻¹⁹)² / (3.2 × 10⁻¹⁵)²
= 90 N

Coulombic Force between them is 90 N

Answer 2

The Coulomb force is 90 N.

Given:

Distance = r = $3.2 \times 10^{-15} \ \mathrm{m}$

To find:

Coulomb force = ?

Solution:

Take,

$\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \ N m^{2} C^{-2}$

Charge on the alpha particle = +2e = $+2 \times\left(1.6 \times 10^{-19}\right) \ C$

Required repulsive Coulomb force between the alpha particles is,

$F=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{q_{1} \times q_{2}}{r^{2}}$

The Coulomb force is directly proportional to the charges and inversely proportional to distance between two charges. This is called as Coulomb’s law.

We know that,

$r=3.2 \times 10^{-15} \ \mathrm{m}$

$\Rightarrow F=\frac{9 \times 10^{9} \times 2 \times 1.6 \times 10^{-19} \times 2 \times 1.6 \times 10^{-19}}{3.2 \times 10^{-15} \times 3.2 \times 10^{-15}} \ N$

$F=\frac{9 \times 10^{9} \times 3.2 \times 3.2 \times 10^{-38}}{3.2 \times 3.2 \times 10^{-30}}$

$\therefore F=90 \ N$