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NEET PHYSICSMedium

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}.

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