Question

You have two metal spheres, one with twice the diameter than the other. The smaller sphere initially has +8.0 nC of charge, while the larger one is uncharged. The two spheres are then connected together by a long, thin wire. What are the final charges on each sphere? Justify in words each relationship you use to solve this problem.

Answer 1

Let the smaller sphere radius be and Larger sphere radius be   
$\Rightarrow R=2r$

Given that the smaller sphere was given a charge of $Q_o=8(10^{-9})C$
and the larger sphere is uncharged
therefore the electric potential on the larger sphere is ZERO while on the smaller sphere is $V=\frac{1}{4\pi\varepsilon _o}\frac{Q_o}{r}$

Now both the spheres are connected with a long thin wire, Therefore the charge flows from the smaller sphere on to the Larger sphere due to the potential difference .the charge flow takes place until the potential on both the spheres become equal

let 'q' amiunt of charge be transfered to larger sphere ,then
Potential on
Larger sphere will be $V_L=\frac{1}{4\pi\varepsilon _o}\frac{q}{2r}$
smaller sphere will be $V_s=\frac{1}{4\pi\varepsilon _o}\frac{Q_o-q}{r}$

but $V_L=V_s$

$\Rightarrow \frac{1}{4\pi\varepsilon _o}\frac{Q_o-q}{r}=\frac{1}{4\pi\varepsilon _o}\frac{q}{2r}$

$\Rightarrow\frac{Q_o-q}{1}=\frac{q}{2}$

$\therefore q=\frac{2Q_o}{3}=5.33(10^{-9})C$
therefore charge on larger sphere will be $q=\frac{2Q_o}{3}=5.33(10^{-9})C$

charge on smaller sphere will be $Q_o-q=\frac{Q_o}{3}=2.67(10^{-9})C$