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 :

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) ( (2r/3) )  (B) ( (1/√2) )  (C) ((r/√[3]2)) (D) ( (2r/√3) )

Tardigrade Admin · Accepted Answer

tanθ = (F/mg) ⇒ (r/2/y) = (kq2/r2mg) ⇒ y propto r3 Therefore ((r'/r))3 = - (y/2/y) ⇒ r' ((1/2))1/3

Q. 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 :

$(\frac{2 r}{3})$

$(\frac{1}{2})$

$(\frac{r}{3 2})$

$(\frac{2 r}{3})$

tan $θ = \frac{F}{m g}$
$\Rightarrow \frac{r /2}{y} = \frac{k q ^{2}}{r ^{2} m g} \Rightarrow y \propto r_{3}$
Therefore $(\frac{r ^{'}}{r})^{3} = - \frac{y /2}{y} \Rightarrow r^{'} (\frac{1}{2})^{1/3}$

Q. 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 :

(32r​)

(2​1​)

(32​r​)

(3​2r​)

tanθ=mgF​ ⇒yr/2​=r2mgkq2​⇒y∝r3​ Therefore (rr′​)3=−yy/2​⇒r′(21​)1/3

Q. 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 :

$(\frac{2 r}{3})$

$(\frac{1}{2})$

$(\frac{r}{3 2})$

$(\frac{2 r}{3})$

tan $θ = \frac{F}{m g}$
$\Rightarrow \frac{r /2}{y} = \frac{k q ^{2}}{r ^{2} m g} \Rightarrow y \propto r_{3}$
Therefore $(\frac{r ^{'}}{r})^{3} = - \frac{y /2}{y} \Rightarrow r^{'} (\frac{1}{2})^{1/3}$