Question

The electrostatic force of repulsion between two positively charged ions carrying equal charges = 3.7 x 10-9n when they are separated by a distance of 5 å. how many electrons are missing from each ion? [how do u tell missing of electrons in this explain​

Answer 1

Given, the force of repulsion between two equally positively charged ions =3.7×10

−19

N

Distance of separation, d=5Å =5×10

10

m

To find the number of electrons missing from each ion.

Let us suppose charge is q

1

=q

2

=q

Let n be the number of electrons missing.

Using Coulomb's Law, F=

4πε

0

1

d

2

q

1

q

2

F=

4πε

0

1

d

2

q

2

So, q

2

=4πε

0

×F×d

2

=

9×10

9

1

×3.7×10

−9

×(5×10

−10

)

2

=10.28×10

−38

q=3.2×10

−19

C

As, q=ne

So, n=

e

q

=

1.6×10

−19

3.2×10

−19

n=2