Question

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

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

Answer 1

The correct option is B $3\times {10}^{-2}C$


So given that
Charge on A $Q_2$ $-1\times {10}^{-2}C$
Charge on B $Q_2=5\times {10}^{-2}$
radius of A $r_1=1cm$
radius of B $r_2=5cm$
Let ${}^{\prime }{q}^{\prime }$ charge gets transfered from A to B through the conducting wire.
Charge on A $=Q_1-q$
Charge on B $=Q_2+q$
And now potential of both spheres become equal
$\frac{k(Q_1-q)}{r_1}=\frac{k(Q_2+q)}{r_2}$
$\Rightarrow \frac{-1-q}{1}=\frac{5+1}{3}\Rightarrow q=-2\times {10}^{-2}$
Now charge on bigger sphere B is given by
$Q=Q_2+q\Rightarrow Q=5-2=3\times {10}^{-2}=3\times {10}^{-2}C$