Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes [figure]
- ( 2r 3 )
- ( 2r3 )
- ( 1 2 )2 r
- ( r [3]2 )
Answer
(4) ( r [3]2 )
Solution
Let θ be the angle of each string with the vertical. For equilibrium of each ball:
T = mg and T = F = kq2r2
⇒ = Fmg
From the figure, = r/2y
⇒ r2y = kq2r2 mg
⇒ r3 = 2kq2 ymg, i.e. r3 y
When the height becomes y2:
r'3r3 = y/2y = 12
∴ r' = r [3]2
Answer: (4)