Question

9 Calaulate the force between two a- particles
seperated by distence of 32 x 10^15 meter​

Answer 1

The Coulomb force is 90 N.

Answer 2

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Take $\frac{1}{4\pi \:Σ \: 0} = 9 \times 10 {}^{9} NM {}^{2} C {}^{ - 2}$

Charge on the Alpha particle = +2E

$= (+ 2)\times 1.6 \times 10 {}^{ - 19 \: } c$

$R = 3.2 \times 10 {}^{ - 15} M$

require repulsive Coulomb force between the alpha particle is,

$F = \frac{1}{4\pi \: Σ \: 0} \times \frac{q {}^{1} \times q {}^{2} }{r {}^{2} } \\ \\ F = \frac{9 \times 10 {}^{9} \times 3.2 \times 3.2 \times 10 {}^{ - 38} }{3.2 \times 3.2 \times 10 {}^{ - 30} } n \\ \\ F = \frac{9 \times 10 {}^{9} \times 3.2 \times 3.2 \times 10 {}^{ - 38} }{3.2 \times 3.2 \times 10 {}^{ - 30} }$

$F = 90N$