Two identical small balls having charges q1
+1mC and q2
= – 0.33 mC are brought in
contact and them moved apart to a distance r
= 20 cm. Find the force of their interaction.
(a) 50 kN (b) 25 kN
(c) 20 kN (d) 10 kN
Answers
Answered by
0
Explanation:
Answer
In the 1st case we have:
F
1
=
d
2
kQ
1
Q
2
(By coulomb's law, k=
4πϵ
0
1
=9×10
9
Nm
2
/C
2
)
After touching the spheres the charges distribute such that charge on sphere is:
q=
2
Q
1
+Q
2
≈
2
Q
1
(Q
1
>>Q
2
)
So new force between them will be:
F
2
=
d
2
kq
2
=
4d
2
kQ
1
2
Hence
F
2
F
1
=
Q
1
2
4Q
1
Q
2
=
Q ni14Q2
.
Similar questions