First slide

Coulumbs law

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:

Difficult

A

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

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

${(\frac{1}{\sqrt{2}})}^{2}$
D

$(\frac{r}{\sqrt[3]{2}})$

Solution

Let m be mass of each ball and q be charge on each ball.
Force of repulsion, $F = \frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}$
$I n e q u i l i b r i u m T \cos θ = m g . . . . . . . . . (i)$
$T \sin θ = F . . . . . . . . (i i)$
Divide (ii) by (i), we get
$\tan θ = \frac{F}{m g} = \frac{\frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}}{m g}$
From figure (a),
$\frac{r / 2}{y} = \frac{\frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}}{m g} - - - (i i i)$
$\tan θ^{'} = \frac{\frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{2}}}{m g}$
From figure (b)
$\frac{r^{'} / 2}{y / 2} = \frac{\frac{1}{4 π ε_{0}} \frac{q^{2}}{r^{' 2}}}{m g} - - - (i v)$
Divide (iv) by (iii), we get
$\frac{2 r^{'}}{r} = \frac{r^{2}}{{r^{'}}^{2}} {r^{'}}^{3} = \frac{r^{3}}{2} r^{'} = \frac{r}{\sqrt[3]{2}}$

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