Advertisements
Advertisements
Question
Two identical conducting spheres A and B, carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B would be equal to ______.
Options
`(3"F")/4`
`"F"/2`
F
`(3"F")/8`
Solution
Two identical conducting spheres A and B, carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B would be equal to `bb((3"F")/8)`.
Explanation:
Say "q" when spheres A and B both have an equal charge.
∴ Force between them, F = `"kqq"/"r"^2`
Charge on both when A and C come into contact qA = qC = `"q"/2`
Charge on B after B and C have made contact.
qB = `("q"/2+"q")/2 = (3"q")/4`
The force between charges qA and qB is now.
F' = `("kq"_"A""q"_"B")/("r"^2)`
= `("k"xx"q"/2xx(3"q")/4)/"r"^2`
= `3/8"kq"^2/"r"^2`
= `3/8"F"`
