find the minimum force between two charge particles kept at separation of 1m
Answers
These two charges are separated by a distance
r
. Then according to Coulomb's Law the force
F
of attraction or repulsion between them is,
F
∝
q
1
q
2
and
F
∝
1
r
2
Therefore,
F
∝
q
1
q
2
r
2
(1)
or,
F
=
k
q
1
q
2
r
2
where,
k
is the constant of proportionality. The value of
k
depends on the nature of the medium between two charges and the system of units we choose to measure
F
,
q
1
,
q
2
and
r
.
In SI units system, when two charges are in vacuum or air
k
=
1
4
π
ε
0
where
ε
0
is absolute permitivity of free space. Value of this constant in vacuum is
8.85
×
10
−
12
C
2
/
N
m
2
. If we put the value of
ε
0
in above equation for
k
we find
k
=
1
4
π
ε
0
=
1
4
×
π
×
8.85
×
10
−
12
C
2
/
N
m
2
=
9
×
10
9
N
m
2
C
−
2
So, from above equation (1) the force between two charges located in air or vacuum is given by,
F
=
1
4
π
ε
0
q
1
q
2
r
2
=
9
×
10
9
×
q
1
q
2
r
2
(in Newton)
Now if the charges are in a medium (glass, water etc.) other then air and vacuum then electric force between these two charges is
F
=
1
4
π
ε
q
1
q
2
r
2
where,
ε
is a constant and is the the permitivity of the medium in which charges are present. Since the value of
ε
depends on the medium, the magnitude of force on a charge also depends on the medium.
Answer:
Explanation:
Answer : (a) 2.3×10−28N
Explanation :
F=14πε0q1q2r2
F will be minimum when q1andq2 are minimum i.e.q1=q2=1.6×10−19
Also, given r=1m.
Fmin=9×109N×(1.6×10−10)2
Fmin=2.3×10−28N.