Given 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, we have to find the equilibrium separation when the strings are clamped at half the height. We can see from the figure that under equilibrium, tanθ=mgF...(i) and also, tanθ=yr/2… (ii) Equating (i) and (ii), we get mgF=2yr also, Force, F=r2kq2 where k= constant =9×109 So, we get r3=(mgkq2)y… (iii) Let r′ be the equilibrium separation when y→2y So, equation (iii) then reduces to: r′3=(mgkq2)2y… (iv) Dividing (iv) by (iii) we get, So, r3r′3=21 r′=r(21)1/3 
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…
NEET UG 2013 — Physics Electromagnetism
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 become. 
Held on 30 Apr 2013 · Verified 9 Jul 2026.
Options
- A
(21)2
- B
(32r)
- C
(32r)
- D
(32r)
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