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NEET PHYSICSELECTRIC CHARGES AND FIELDSMedium

Question

Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r (as shown in Fig. I). Now, as shown in Fig. II, the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes:

A

r / ∛2

B

r / √2

C

2r / 3

D

none of the above

Step-by-Step Solution

At equilibrium, the electrostatic repulsive force (FeF_e) is balanced by the component of gravity acting along the tangent of the arc. For small angles θ\theta, the condition for equilibrium is tanθ=Femg\tan \theta = \frac{F_e}{mg}.

  1. Geometry: From the figure, tanθsinθ=r/2y=r2y\tan \theta \approx \sin \theta = \frac{r/2}{y} = \frac{r}{2y}, where yy is the vertical distance from the point of suspension to the line joining the balls.
  2. Force Equation: Substituting Coulomb's law (Fe=kq2r2F_e = \frac{kq^2}{r^2}), we get kq2r2mg=r2y\frac{kq^2}{r^2 mg} = \frac{r}{2y}.
  3. Proportionality: Rearranging the terms, r3yr^3 \propto y.
  4. New Condition: When clamped at half height, the new vertical distance is y=y/2y' = y/2. Let the new separation be rr'.
  5. Calculation: (r)3r3=yy=12    r=r2\frac{(r')^3}{r^3} = \frac{y'}{y} = \frac{1}{2} \implies r' = \frac{r}{\sqrt{2}}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTRIC CHARGES AND FIELDS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSELECTRIC CHARGES AND FIELDScarryingchargessuspendedcommonstrings

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