Question:

Two balls of same mass and carrying equal charge are hung from a fixed support of length $l$. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, $x$ between the balls is proportional to :

Updated On: Aug 14, 2024
  • $l$
  • $l^2$
  • $l^{2/3}$
  • $l^{1/3}$
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The Correct Option is D

Solution and Explanation

In equilibrium, $F_{e} = T \,sin\, \theta$ $mg = T\,cos\,\theta$ $tan\,\theta = \frac{F_{e}}{mg} = \frac{q^{2}}{4\pi\,\in_{0}\,x^{2}\times mg}$ also $tan\, \theta\,\approx\, sin = \frac{x/2}{\ell}$ Hence, $ \frac{x}{2\ell } = \frac{q^{2}}{4\pi \,\in _{0}\,x^{2}\times mg}$ $\Rightarrow x^{3} = \frac{2q^{2}\ell}{4\pi \,\in_{0}\,\times mg}$ $\therefore x = \left(\frac{q^{2}\ell }{2\pi \,\in _{0}\,\times mg}\right)^{1/3}$ Therefore $x \,\propto \,\ell^{1/3}$
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Concepts Used:

Electric Field

Electric Field is the electric force experienced by a unit charge. 

The electric force is calculated using the coulomb's law, whose formula is:

\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)

While substituting q2 as 1, electric field becomes:

 \(E=k\dfrac{|q_{1}|}{r^{2}}\)

SI unit of Electric Field is V/m (Volt per meter).