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Two charge particles are placed at 1 m apart what is the minimum possible magnitude of electric force acting on each charge

$(a)2.3\times10^{-28}\;N\;\qquad(b)\;2.3\times10^{-8}\;N\qquad(c)\;2.3\times10^{-24}\;N\qquad(d)\;2.3\times10^{-6}\;N$

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Answer : (a) $\;2.3\times10^{-28}N$
Explanation :
$\;F= \large\frac{1}{4\pi\varepsilon_0} \frac{q_{1}q_{2}}{r^2}$
F will be minimum when $\;q_{1}\;and\;q_{2}\;$ are minimum $\;i .e . \;q_{1}=q_{2}=1.6\times10^{-19}$
Also, given $r = 1m$.
$F_{min}= 9\times10^9\;N \times (1.6 \times 10^{-10})^2 $
$F_{min}=2.3\times10^{-28}\;N\;.$
answered Jan 29, 2014 by yamini.v
edited Mar 23, 2014 by balaji.thirumalai
 

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