Question

Five electrons have been removed from each of two atoms to form ions. These ions are placed in free
space with a distance 20 A between them. Find the electrostatic force between them.​

Answer 1

Answer:

F=K(Q/(2×10-9)-4×10-10) WHERE Q=TOTAL CHARGE OF ATOM BEFORE REMOVING OF ELECTRON.

F=K × q1 q2 /r^2

q1=q2=q,q=ne. .....Q=5×1.6×10^-19=8×10^-19

F=9×10^9×8×10^-19×8×10^-19/20×20×10^-10×10^-10=1.44nN

Answer 2

The electrostatic force between them is $1.44\times 10^{-9}\ N$.

Explanation:

It is given that, five electrons have been removed from each of two atoms to form ions. Now, charge on both ions will be :

$q_1=q_2=5e=5\times 1.6\times 10^{-19}=8\times 10^{-19}\ C$

Distance between charges, $d=20\ A=20\times 10^{-10}\ m$

The electrostatic force between them is given by :

$F=\dfrac{1}{4\pi \epsilon_o}\dfrac{q_1q_2}{d^2}\\\\F=9\times 10^9\times \dfrac{(8\times 10^{-19})^2}{(20\times 10^{-10})^2}\\\\F=1.44\times 10^{-9}\ N$

So, the electrostatic force between them is $1.44\times 10^{-9}\ N$.

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