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Q.
Two metal spheres of radii $0.01 \,m$ and $0.02\, m$ are given a charge of $15\, mC$ and $45\, mC$ respectively. They are then connected by a wire. The final charge on the first sphere is _____ $\times 10^{-3}C$
KCETKCET 2013Electrostatic Potential and Capacitance
Solution:
When two metal spheres are joined by a wire, charge flows from one at higher potential to the other at lower potential, till they acquire the same potential.
$\therefore V_{1}=V_{2}$
$\frac{q_{1}}{4 \pi \varepsilon_{0} r_{1}} =\frac{q_{2}}{4 \pi \varepsilon_{0} r_{2}} $
$ \Rightarrow \frac{r_{1}}{r_{2}}=\frac{q_{1}}{q_{2}} =\frac{15\,m C}{45\, m C}=\frac{1}{3}$
Final charge on first sphere,
$q_{1}=\frac{1}{3}\left(q_{1}+q_{2}\right)=\frac{1}{3} \times 60=20\, mC$