Question

What is the value of minimum force acting between two charges placed at 1m?

Answer 1

Answer:

We know that, F=(1/4πε0)(q^1q^2)/r^2. F will be minimum when q1andq2 are minimum i. e. q1=q2=1.6×10−19.2

Answer 2

Answer:

Minimum electric charge will be = 1.6 X $10^{-19}$C

So, Minimum force of attraction = $\frac{k q^{2} }{r^{2} }$ = $\frac{9 X 10^{9} X 1.6 X 10^{-19} X 1.6 X 10^{-19} }{1^{2} }$

  = 23.04 X $10^{-29}$ = 2.304 X $10^{-28}$N