From.
Unit 6
Electrostatics
Q.45 The ratio of the forces between two small conducting spheres charged to constant
potentials in (a) air (b) a medium of K = 2 is
C) 1:2
D) 4:1
A) 1:4
B) 2:1
Answers
Answered by
0
Answer:
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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