The distance between electron and proton in the H-atom is about 5-3 - 10-11 metre- What is the magnitude of electric force between these two particles- 14 -0 - 9 - 109 Nm2C2

The electric #10;force between the particles is given as, #10; #10; F = 14π #10;ε _0×Q_1Q_2r^2 #10; #10; F = 9 ×10^9 #10;×1.6 ×10^ 19× ...

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Standard XII

Physics

Coulomb's Law - Mathematical Form

The distance ...

Question

The distance between electron and proton in the H-atom is about $5.3 \times 10^{- 11}$ metre. What is the magnitude of electric force between these two particles?
$(\frac{1}{4 π ϵ_{0}} = 9 \times 10^{9} \frac{N m^{2}}{C^{2}})$

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Solution

The electric force between the particles is given as,

$F = \frac{1}{4 π ε_{0}} \times \frac{Q_{1} Q_{2}}{r^{2}}$

$F = 9 \times 10^{9} \times \frac{1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{{(5.3 \times 10^{- 11})}^{2}}$

$F = 0.82 \times 10^{- 7} N$

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The distance between electron and proton in the H-atom is about 5.3×10−11 metre. What is the magnitude of electric force between these two particles?(14πϵ0=9×109Nm2C2)

The electric force between the particles is given as, F=14πε0×Q1Q2r2 F=9×109×1.6×10−19×1.6×10−19(5.3×10−11)2 F=0.82×10−7N

The distance between electron and proton in the H-atom is about $5.3 \times 10^{- 11}$ metre. What is the magnitude of electric force between these two particles?
$(\frac{1}{4 π ϵ_{0}} = 9 \times 10^{9} \frac{N m^{2}}{C^{2}})$

The electric force between the particles is given as,

$F = \frac{1}{4 π ε_{0}} \times \frac{Q_{1} Q_{2}}{r^{2}}$

$F = 9 \times 10^{9} \times \frac{1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{{(5.3 \times 10^{- 11})}^{2}}$

$F = 0.82 \times 10^{- 7} N$