Question

The electrostatic force of repulsion between two positive ions carrying equal charges is 4 x10^-9 N, when their separation is 5 A°(angstrom). How many electrons are missing from each ?​

Answer 1

Answer:

Explanation:

Let's assume the charges to be +q.

F = kq2r2

k = 9×109Nm2C−2,

F=4×10−9N,

r=5×10−10m

⟹4×10−9 = 9×109.q225×10−20

Or, q = 3.3×10−19C

q = -ne

⟹n=−3.3×10−19−1.6×10−19

No. Of electrons missing in each, n= 2

Answer 2

Number of electrons are 2.5

Explanation:

Given that,

Electrostatic force between two positive ions, $F=4\times 10^{-9}\ N$

Separation between the ions, $d=6A=6\times 10^{-10}\ m$

We know that the electrostatic force of repulsion between two positive ions is given by :

$F=k\dfrac{q^2}{d^2}$

$q=\sqrt{\dfrac{Fd^2}{k}}$

$q=\sqrt{\dfrac{4\times 10^{-9}\times (6\times 10^{-10})^2}{9\times 10^9}}$

$q=4\times 10^{-19}\ C$

Using the quantization of charges :

q = ne

n is the number of electrons

$n=\dfrac{q}{e}$

$n=\dfrac{4\times 10^{-19}}{1.6\times 10^{-19}}$

n = 2.5 electrons

Learn more :

Topic : Electrostatic force

https://brainly.in/question/10522197

https://brainly.in/question/3749896