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Physics Question on Electrostatic potential

Two metallic spheres of radii 1 cm and 3 cm are are given charges of −1×10−2C-1 \times 10^{-2} C and 5×10−2C5 \times 10^{-2} C , respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is

A

2×10−2C2 \times 10^{-2} C

B

3×10−2C3 \times 10^{-2} C

C

4×10−2C4 \times 10^{-2} C

D

1×10−2C1 \times 10^{-2} C

Answer

3×10−2C3 \times 10^{-2} C

Explanation

Solution

When the given metallic spheres are connected by a conducting wire, charge will flow till both the spheres acquire a common potential which is given by
Common potential,
V=q1+q2C1+C2=−1×10−2+5×10−24πε0R1+4πε0R2V = \frac{q_1 + q_2}{C_1 + C_2} = \frac{-1 \times 10^{-2} + 5 \times 10^{-2}}{4\pi \varepsilon_0 R_1 + 4 \pi \varepsilon_0 R_2}
=4 × 10−24πε0(1 × 10−2+3 × 10−2)= \frac{4\, \times \,10^{-2}}{4\pi \varepsilon_0 (1 \,\times\,10^{-2}+3 \,\times\,10^{-2})}
=4 × 10−24πε0 × 4 × 10−2   ...(i)= \frac{4 \,\times\, 10^{-2}}{4\pi \varepsilon_0\, \times\, 4\, \times \,10^{-2}}\, \, \, ...(i)
∴\therefore Final charge on the bigger sphere is
=4πε0×3×10−2×4 × 10−24πε0 × 4 × 10−2= 4\pi \varepsilon _0\times 3 \times 10^{-2}\times\frac{4\, \times \,10^{-2}}{4\pi \varepsilon_0\, \times \,4 \, \times \,10^{-2}} (Using (i))
=3×10−2C= 3 \times 10^{-2} C