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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^{-2}coulomb ;

V_{1} = V_{2}

$\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