Question

If two unit positive charges are separated by a
distance of 1 m,then the electrostatic force of
repulsion between them is...
A) 1000N
B) 1N
C) 9 x 10 to the power 9 N
D) 60x10to the power 9 NI​

Answer 1

Answer:

2

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