Question

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:





A)




B)




C)




D)




explain properly.......

Answer 1

Let tension in string is T.

horizontal component of tension is balanced by electrostatic force of repulsion.

or, $Tsin\theta=\frac{kq^2}{r^2}$....(1)

and vertical component of tension is balanced by weight of ball.

or, $Tcos\theta=mg$......(2)

from equations (1) and (2),

$tan\theta=\frac{kq^2}{r^2mg}$

from diagram, $tan\theta=\frac{r/2}{y}$

so, $\frac{r/2}{y}=\frac{kq^2}{r^2mg}$

or, $y=\frac{r^3mg}{2kq^2}$

here it is clear that y is directly proportional to r³

or, r is directly proportional to y⅓

so, $\frac{r'}{r}=\left(\frac{y/2}{y}\right)^{1/3}$

or, $r'=r\left(\frac{1}{2}\right)^{1/3}$

hence, answer is $r'=r\left(\frac{1}{2}\right)^{1/3}$