If two unit positive charges are separated by adistance of 1 m,then the electrostatic force ofrepulsion between them is…A) 1000NB) 1NC) 9 x 10 to the power 9 ND) 60x10to the power 9 NI About the author Samantha
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 Reply
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