The electrostatic force on a small sphere of charge 0.2 µC due to another small sphere of charge -0.4 µC in air is 0.4 N. The distance between the two spheres is?
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a)
Electrostatic force on the first sphere, F=0.2N
Charge on this sphere, q
1
=0.4μC=0.4×10
−6
C
Charge on the second sphere, q
2
=−0.8μC=−0.8×10
−6
C
Electrostatic force between the spheres is given by the relation,
F=
4πε
0
r
2
q
1
q
2
And,
4πε
0
1
=9×10
9
Nm
2
C
−2
Where, ε
0
= Permittivity of free space
r
2
=
4πε
0
F
q
1
q
2
=
0.2
04×10
−6
×8×10
−6
×9×10
9
=144×10
−4
r=
144×10
−4
=0.12m
The distance between the two spheres is 0.12 m.
(b) Equal and Opposite Force acts on the other sphere (By newton's Third Law). Hence 0.2 N attractive.
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