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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

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 $

$F_{min}=2.3\times10^{-28}\;N\;.$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...