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  • Please,please help me Two pith balls carrying equal charges aresuspended from a common point by strings of equallength, the equilibrium separation between them isr. Now the strings are rigidly clamped at half theheight. The equilibrium separation bet

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:

 

  • Option 1)

    \left ( \frac{2\text{r}}{3} \right )

  • Option 2)

    \left ( \frac{1}{\sqrt{2}} \right )^{2}

  • Option 3)

    \left ( \frac{\text{r}}{\sqrt[3]{2}} \right )

  • Option 4)

    \left ( \frac{2\text{r}}{\sqrt{3}} \right )

 

Answers (1)

best_answer

As discussed in @7916

F = \frac{1}{4\pi s_{0}}\:\frac{q^{2}}{r^{2}}

T\cos \Theta = mg\:\:\:-i

T\sin \Theta = F\:\:\:-ii

\therefore from equation (i) and (ii)

\tan \Theta = \frac{f}{mg}= \frac{\frac{1}{4\pi \varepsilon _{0}}\frac{q^{2}}{r^{2}}}{mg}

\frac{\frac{r}{2}}{Y}= \frac{\frac{1}{4\pi \varepsilon _{0}}\frac{q^{2}}{r^{2}}}{mg}\:\:\:-iii

From figure (b)

\frac{r\frac{1}{2}}{\frac{y}{2}} = \frac{\frac{1}{4\pi \varepsilon _{0}}\frac{q^{2}}{r'^{2}}}{mg}\:\:\:-iv

\therefore from equation (iii) and (iv)

\frac{2r'}{r}= \frac{r^{2}}{r'^{2}}\Rightarrow r'^{3} = \frac{r^{3}}{2}\Rightarrow r' = \frac{r}{\sqrt[3]{2}}


Option 1)

\left ( \frac{2\text{r}}{3} \right )

This answer is incorrect

Option 2)

\left ( \frac{1}{\sqrt{2}} \right )^{2}

This answer is incorrect

Option 3)

\left ( \frac{\text{r}}{\sqrt[3]{2}} \right )

This answer is correct

Option 4)

\left ( \frac{2\text{r}}{\sqrt{3}} \right )

This answer is incorrect

Posted by

Aadil

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