Question

Calculate the force of attraction between an electron and a proton, seperated at the distance of 10 Á​

Answer 1

Explanation:
From Coulomb's inverse square law: F = $\frac{1}{4\piε_{0} } \frac{q1q2}{r\\{2} }$$\frac{1}{4\piε_{0} } \frac{q1q2}{r^{2}}$
The magnitude of the charge on electron and proton is the same =$1.6*10^{-19} C$
Therefore, an electrostatic force is given by
$F=\frac{9*10^{9}*1.6*10^{-19}*1.6*10^{-19} }{(0.05*10^{-9})^{2} }$
=$\frac{9*1.6*1.6*10^{-29} }{(5*10^{-11)}^{2} }$
$=\frac{9*1.6*1.6*10^{-29} }{25*10^{-22} }$
$=0.9216*10^{-29} *10^{+22}$
$=0.9216*10^{-7}N$
=  $9.216*10^{-8} N$

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