Two small equal point charges of magnitude q are suspended from a common point on the ceiling by insulating mass less strings of equal lengths. They come to equilibrium with each string making angle θ from the vertical. If the mass of each charge is m, then the electrostatic potential at the centre of line joining them will be ((1)/(4πε₀) = k). (a) 2√(kmgtan θ) (b) √(kmgtan θ) (c) 4√(kmgtan θ) (d) 4√(kmg/tan θ)

Answer: (c). Equilibrium of each charge: (kq²)/(r²) = mgtan θ ⇒ q/r = √((mgtan θ)/(k)). At the midpoint each charge is at r/2: V = 2 × kq/r/2 = 4kq/r. V = 4k√((mgtan θ)/(k)) = 4√(kmgtan θ) So the answer is (c).

Two small equal point charges of magnitude q are suspended from a… | JEE Main 2013 (22 Apr, Shift 1)

Q71JEE Main 2013 · 22 APR · SHIFT 1Medium

Electric Charge & Coulomb's Law

Two small equal point charges of magnitude $q$ are suspended from a common point on the ceiling by insulating mass less strings of equal lengths. They come to equilibrium with each string making angle $\theta$ from the vertical. If the mass of each charge is $m$ , then the electrostatic potential at the centre of line joining them will be $\left(\frac{1}{4\pi\epsilon_0} = k\right)$ .

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Answer: (c)

Equilibrium of each charge: $\dfrac{kq^2}{r^2} = mg\tan\theta \Rightarrow \dfrac{q}{r} = \sqrt{\dfrac{mg\tan\theta}{k}}$ .
At the midpoint each charge is at $r/2$ : $V = 2 \times \dfrac{kq}{r/2} = \dfrac{4kq}{r}$ .
$V = 4k\sqrt{\dfrac{mg\tan\theta}{k}} = 4\sqrt{kmg\tan\theta}$
So the answer is (c).

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