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NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEMedium

Question

Two metallic spheres of radii 1 cm and 3 cm are given charges of -1×10⁻² C and 5×10⁻² C, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is:

A

2×10⁻² C

B

3×10⁻² C

C

4×10⁻² C

D

1×10⁻² C

Step-by-Step Solution

  1. Conservation of Charge: When the two spheres are connected, the total charge is conserved. Qtotal=Q1+Q2=(−1×10−2)+(5×10−2)=4×10−2 CQ_{total} = Q_1 + Q_2 = (-1 \times 10^{-2}) + (5 \times 10^{-2}) = 4 \times 10^{-2} \text{ C}.

  2. Common Potential: Charge flows until the electric potential is the same on both spheres (V1=V2V_1 = V_2). For a spherical conductor, Potential V=kQ/RV = kQ/R. Thus, Q1′/R1=Q2′/R2Q'_1/R_1 = Q'_2/R_2, which implies that the final charge is proportional to the radius (Q∝RQ \propto R).

  3. Redistribution Formula: The final charge on a sphere is a fraction of the total charge based on its capacitance (or radius). Qbigger′=Qtotal×RbiggerRsmall+RbiggerQ'_{bigger} = Q_{total} \times \frac{R_{bigger}}{R_{small} + R_{bigger}}

  4. Calculation: Given R1=1 cmR_1 = 1 \text{ cm}, R2=3 cmR_2 = 3 \text{ cm} (bigger sphere). Qbigger′=(4×10−2)×31+3Q'_{bigger} = (4 \times 10^{-2}) \times \frac{3}{1 + 3} Qbigger′=(4×10−2)×34=3×10−2 CQ'_{bigger} = (4 \times 10^{-2}) \times \frac{3}{4} = 3 \times 10^{-2} \text{ C}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROSTATIC POTENTIAL AND CAPACITANCE. 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.

PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEmetallicsphereschargesrespectivelyconnected

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