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Physics Question on electrostatic potential and capacitance

Two metallic spheres of radii 1 cm and 2 cm are given charges 10−2C 10^{ - 2} C and 5×10−2C5 \times 10^{ - 2} C respectively. If they are connected by a conducting wire, the final charge on the smaller sphere is

A

3×10−2 C3 \times 10^{ - 2} \, C

B

4×10−2 C4 \times 10^{ - 2} \,C

C

1×10−2 C1 \times 10^{ - 2} \,C

D

2×10−2 C2 \times 10^{ - 2} \,C

Answer

2×10−2 C2 \times 10^{ - 2} \,C

Explanation

Solution

The correct answer is D:2×10−2 C2 \times 10^{ - 2} \,C
Radii of sphere (R1)=1(R_1) = 1 cm = 1×10−21 \times 10^{ - 2} m ;
(R2)=2 cm=2×10−2(R_2) = 2 \, cm = 2 \times 10^{ - 2} m and charges on sphere;
(Q1)=10−2C(Q_1) = 10^{ - 2} C and (Q2)=5×10−2(Q_2) = 5 \times 10^{ - 2} C.
Common potential (V)
=Total chargeTotal capacity= \frac{ Total \, charge }{ Total \, capacity}
=Q1+Q2C1+C2=(1×10−2)+(5×10−2)4πε010−2+4πε0(2×10−2)= \frac{ Q_1 + Q_2 }{ C_1 + C_2} = \frac{( 1 \times 10^{ - 2} ) + ( 5 \times 10^{ - 2}) }{ 4 \pi \varepsilon_0 10^{ - 2} + 4 \pi \varepsilon_0 (2 \times 10^{ - 2} )}
= 6×10−24πε0(3×10−2)\frac{ 6 \times 10^{ - 2}} { 4 \pi \varepsilon_0 (3 \times 10^{ - 2}) }
Therefore final charge on smaller sphere (C1V)(C_1 V)
= 4πε0×10−2×6×10−24πε0×3×10−2=2×10−24 \pi \varepsilon_0 \times 10^{ - 2} \times \frac{ 6 \times 10^{ - 2}} { 4 \pi \varepsilon_0 \times 3 \times 10^{ - 2} } = 2 \times 10^{ - 2} C.