Question

Two identical pith balls, each carrying a charge q, are suspended from a common point by two strings of equal length l. Find the mass of each ball if the angle between the strings is 2θ in equilibrium.

Answer 1

Explanation:

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Answer 2

The mass of each ball if the angle between the strings is 2θ in equilibrium is given by $\begin{equation}m=\frac{q^{2} \cot \theta}{16 \pi \epsilon_{0} g l^{2} \sin ^{2} \theta}\end$  .

Explanation:

Given data in the question :

Let m is the mass of each ball and two strings of equal length L are suspended from a common point.

The angle of equilibrium between the stings is 2θ at equilibrium ,

T cos θ = mg    -----> eqn (1)

$\begin{equation}\mathrm{T} \sin \theta=\mathrm{F}\end$   -----> eqn (2)                            

And,                      

      $\begin{equation}\tan \theta=\frac{F}{m g}\end$            

From the above equation, the mass is

$\begin{equation}m=\frac{F}{g \tan \theta}\end$   -----> eqn (3)

Here, separation between the two charges,

$\begin{equation}r=2 l \sin \theta\end$

$\begin{equation}F=\frac{1}{4 \pi \epsilon_{0}} \frac{q^{2}}{r^{2}}\end$

   $\begin{equation}=\frac{1}{4 \pi \epsilon_{0}} \times \frac{q^{2}}{(2 l \sin \theta)^{2}}\end$

Substituting the value of Fin equation (3), we get

$\begin{equation}m=\frac{q^{2} \cot \theta}{16 \pi \epsilon_{0} g l^{2} \sin ^{2} \theta}\end$.

Thus, the mass of each ball is $\begin{equation}m=\frac{q^{2} \cot \theta}{16 \pi \epsilon_{0} g l^{2} \sin ^{2} \theta}\end$.