Two metallic spheres of radii 1 cm and 3 cm are given charges of -1×10^(-2) C and 5×10^(-2) C, respectively….

Q: 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

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Ans: (b)
Sol: When the two spheres are connected by a conducting wire, change flows from second sphere to fist sphere till their potentials becomes equal . Let the bigger sphere losses charge x × 10-2 coulomb ;

V1 = V2

$\displaystyle \frac{1}{4\pi \epsilon_0} \frac{q_1}{r_1} = \frac{1}{4\pi \epsilon_0} \frac{q_2}{r_2} $

$\displaystyle \frac{1}{4\pi \epsilon_0} \frac{x + q_1}{r_1} = \frac{1}{4\pi \epsilon_0} \frac{q_2 – x}{r_2} $

$\displaystyle \frac{(x – 1)\times 10^{-2}}{1 \times 10^{-2}} = \frac{(5 – x)\times 10^{-2}}{3 \times 10^{-2}} $

3 x – 3 = 5 – x

4 x = 8 , x = 2

Charge left on bigger sphere =(5-2)10-2 C

= 3 × 10-2 C