Question

Calculate coulomb force between two alpha particles when at a distance of 10^-13m from eachother charge of an alpha particle is 3.2×10^-12C. If mass of each particle is 6.08×10^-27kg(take k= 9×10^-9 and G=6.67 ×10^-11

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{F_{E}=9.216\times 10^{12}\:N}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Distance \: between \: particles = {10}^{ - 13} \: m \\ \\ \tt: \implies Charge \: of \: particle = 3.2 \times {10}^{ - 12} \: \: \: \: \: \: (each) \\ \\ \tt: \implies Mass \: of \: particle = 6.08 \times {10}^{ - 27} \: kg \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Force \: between \: two \: particles =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies F_{E} = \frac{kq_{1}q_{2} }{ {r}^{2} } \\ \\ \tt: \implies F_{E} = \frac{9 \times {10}^{ 9} \times 3.2 \times {10}^{ - 12} \times 3.2 \times {10}^{ - 12} }{ ({10}^{ - 13})^{2} } \\ \\ \tt: \implies F_{E} = \frac{9 \times 3.2 \times 3.2 \times {10}^{9 - 24} }{ {10}^{ - 26} } \\ \\ \tt: \implies F_{E} =92.16 \times {10}^{ - 15 + 26} \\ \\ \tt: \implies F_{E} =92.16 \times 10^{11} \\ \\ \green{\tt: \implies F_{E} =9.216 \times {10}^{12} \: N}$

Answer 2

Answer:

• Coulomb Force will be 9.216 × 10¹² N

Given:

1. Charge (q) = 3.2 × 10^{- 12} C
2. Separation (r) = 10^{-13} m

Explanation:

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Here, we need to just simply substitute the values in the Coulombs formula.

From the formula we know,

F = K q q / r²

Here,

• F Denotes force.
• K Denotes constant.
• q Denotes charges.
• r Denotes separation.

Now, refer the attachment for calculations.

∴ Coulomb force will be 9.216 × 10¹² N.

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The Force will be repulsive as both have same signs and we know that like charges repel each other.

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