Question

calculate the force between two alpha particles seperated by the distence of 3.2×10^15 meter.​

Answer 1

Answer:

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

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Answer 2

Answer:

The column force is 90N.

Given :

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

To find :

Column force = ?

Solution :

Take,

$\frac{1}{4\pi0} = 9 \times 10 {}^{9 \: nm {}^{2} c {}^{ - 2} }$

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

Required repulsive Coulomb force between the alpha particle is ,

$f \: = \frac{1}{4\pi 0} \times \: \frac{q1 \times q2}{r {}^{2} }$

the Coulomb force us directly proportional to the charges and inversely proportional to distance between two charges . this is called as coulomb's law .

We know that ,

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

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

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

$F\: = \: 90N$