Question

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

• A. $ 3 \times 10^{ - 2} \, C $
• B. $ 4 \times 10^{ - 2} \,C $
• C. $ 1 \times 10^{ - 2} \,C $
• D. $ 2 \times 10^{ - 2} \,C $

Answer 1

Answer: (D)

Radii of sphere $ (R_1) = 1 $ cm = $ 1 \times 10^{ - 2} $ m ;
$ (R_2) = 2 \, cm = 2 \times 10^{ - 2} $ m and charges on sphere;
$ (Q_1) = 10^{ - 2} C$ and $(Q_2) = 5 \times 10^{ - 2} $ C.
Common potential (V)
$ = \frac{ Total \, charge }{ Total \, capacity}$
$ = \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} )} $
= $ \frac{ 6 \times 10^{ - 2}} { 4 \pi \varepsilon_0 (3 \times 10^{ - 2}) } $
Therefore final charge on smaller sphere $ (C_1 V)$ = $ 4 \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.