Solution :
Given :
a)
q 1 = 2 × 10 – 7 C
b)
q 2 = 3 × 10 – 7 C
c)
r = 30
cm = 30 X 10 – 3 = 3 X 10 – 2
d)
ε
0 = 8.854 × 10 – 12 C 2 N –1 m–
2
Formula :
The force between two small charged spheres can be given as
F = [ 1 / ( 4 π ε0 ) ] [ ( q1
q2) / ( r2 ) ]
Therefore,
F = [ 1 / ( 4 X 3.14 X 8.854 × 10–12
) ] { ( 2 × 10–7 X 3 × 10–7) / [ ( 3 X 10– 2)2
] }
F = [ 1/(1.11 × 10–10 )
] [ ( 6 × 10–14 ) / ( 9 X 10 –4 ) ]
F = [ 9 X 10 9 ) ] [ ( 6
× 10–10 ) / ( 9 ) ]
F = 60 N
The force between two small charged spheres having charges
of 2 × 10–7C and 3 × 10–7C placed 30 cm apart in air is 60 newton
