two extremely small charged copper spheres have their centres separated by distance 50 cm in vacuum (a) what is mutual force of repulsion if charge on each is 6.5 × 10^-7 c .(b) what will be the force of repulsion if two spheres are placed in water
Answers
Answer:
Solution
(a)
Charge on sphere A, q
A
=6.5×10
−7
C
Charge on sphere B, q
B
=6.5×10
−7
C
Distance between the spheres, r=50cm=0.5m
Force of repulsion between the two spheres,
F=
4π∈
0
r
2
q
A
q
B
Where, ∈
0
= Free space permittivity
4π∈
0
1
=9×10
9
Nm
2
C
2
∴F=
(0.5)
2
9×10
9
×(6.5×10
−7
)
2
=1.52×10
−2
N
Therefore, the force between the two spheres is 1.52×10
−2
N
(b)
After doubling the charge,
Charge on sphere A, q
A
=2×6.5×10
−7
C=1.3×10
−6
C
Charge on sphere B, q
B
=2×6.5×10
−7
C=1.3×10
−6
C
Now, if the distance between the sphere is halved, then
r=
2
0.5
=0.25m
Force of repulsion between the two sphere,
F=
4π∈
0
r
2
q
A
q
B
=
(0.25)
2
9×10
9
×1.3×10
−6
×1.3×10
−6
=0.24336N
Therefore, the force between the two sphere is approximately 0.243 N.