## Pages

### Question on electric charge

Question
Two isolated metallic spheres of radii R and 2R are charged such that both these have the same charge densiti σ. The sphares are located far away from each other and connected by a thin conducting wire. Find the new charge density on the bigger wire.
Solution
Initial charges on sphere of radius R and 2R respectively are
Q1= surface area × charge density = 4πR2σ
and Q2=4π(2R)2σ = 16πR2σ
therefore Q1+Q2=4πR2σ+16πR2σ (1)
As the charges are different, their initial potentials are also different. When the spheres are connected by long thin conducting wire, the charges are redistributed untill their potentials become equal. Let Q'1 and Q'2 be the new charges on spheres of radii R and 2R respectively, their common potential is
$V=\frac{1}{4\pi {\epsilon }_{0}}.\frac{{Q\text{'}}_{1}}{R}=\frac{1}{4\pi {\epsilon }_{0}}.\frac{{Q\text{'}}_{2}}{2R}$
which gives
${Q\text{'}}_{1}=\frac{{Q\text{'}}_{2}}{2}$
Therefore,
${Q\text{'}}_{1}+{Q\text{'}}_{2}=\frac{3{Q\text{'}}_{2}}{2}$
From Law of conservation of charge
Q1+Q2=Q'1+Q'2
Using equation 1 and 2 we have
${Q\text{'}}_{2}=\frac{40\pi {R}^{2}\sigma }{3}$
The new charge density on the bigger sphere of radius R is
$\sigma \text{'}=\frac{{Q\text{'}}_{2}}{4\pi {\left(2R\right)}^{2}}=\frac{5\sigma }{6}$