Question

Two identical pith balls, each of mass m and charge q are suspended from a common point by two strings of equal length l. Find charge q in terms of given quantities, if angle between the strings is 2theta in equilibrium.

Answer 1

Explanation:

Let m is the mass of each ball and balls are suspended from a common point by two strings of equal length L, at equilibrium condition angle between stings is 2θ as shown in figure.

at equilibrium ,

Tcosθ = mg

And , Tsinθ = Kq²/r² ,where r = 2Lsinθ

∴ Tsinθ/Tcosθ = Kq²/r²mg

⇒tanθ = Kq²/(2Lsinθ)²mg

⇒ m = kq²/{4L²sin²θ × tanθ × g}

= Kq²cosθ/4L²sin³θg

Hence, mass of each ball is kq²cosθ/

4L²sin³θg

Answer 2

Answer:

Given:

Identical pith balls have mass m and charge q. They are suspended from a common point by strings of equal

length l.

To find:

q in terms of other quantities when angle between the strings is 2θ.

Concept:

Since mass and charge are same for both the balls , the angle will be θ in-between the string and perpendicular.

Diagram:

Look at attached photo to understand better.

Calculation:

Let tension in strings be T.

$T \cos( \theta) = mg.......(1)$

$T \sin( \theta) = \dfrac{1}{4\pi \epsilon} \dfrac{ {q}^{2} }{ \{ {2l \sin( \theta) \}}^{2} } .......(2)$

Dividing (2) by (1) :

$\tan( \theta) = \dfrac{\dfrac{1}{4\pi \epsilon} \dfrac{ {q}^{2} }{ \{ {2l \sin( \theta) \}}^{2} }}{mg}$

Let 1/(4πε) be k

$\tan( \theta) = \dfrac{ \dfrac{ k{q}^{2} }{ \{ {2l \sin( \theta) \}}^{2} }}{mg}$

Simplifying this :

$\boxed{ \dfrac{k {q}^{2} }{4 {l}^{2}mg } = \tan( \theta) \times {sin}^{2} ( \theta)}$

So final answer is as :

$\boxed{ \red{ \dfrac{k {q}^{2} }{4 {l}^{2}mg } = \tan( \theta) \times {sin}^{2} ( \theta)}}$