Question

Please solve this:-

Two identical pith balls of mass m and having charge q are suspended from a point by weightless strings of length l. If both strings make an angle of θ with the vertical then the distance between the balls will be ( taking θ to be small ) :-

Answer:-(q^2 x l/2π E0 mg)^1/3

Answer 1

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 ,

downward force = upward force

mg = Tcosθ ......(1)

forward force = backward force

Fe = Tsinθ

or, Kq²/r² = Tsinθ ......(2)

where , r is the distance between balls at equilibrium.

from equation (1) and (2),

Tsinθ/Tcosθ = Kq²/r²mg

⇒r² = Kq²/mgtanθ

from figure, tanθ = (r/2)/l = r/2l and K = 1/4π$\epsilon_0$

⇒ r² = 1/4π$\epsilon_0$ × q²/mg(r/2l)

⇒r³ = q²l/2π$\epsilon_0$mg

so, r = $\left(\frac{q^2l}{2\pi\epsilon_0mg}\right)^{1/3}$. This is required answer .

Answer 2

Answer:Please solve this:-

Two identical pith balls of mass m and having charge q are suspended from a point by weightless strings of length l. If both strings make an angle of θ with the vertical then the distance between the balls will be ( taking θ to be small ) :-

Explanation: