Two conducting spheres of radius a and 2a with same charge density (sigma) are placed far away from each other. When they are connected by a conducting wire then due to charge redistribution their potential changes. Find the common potential of each sphere
Given that the radius of smaller sphere is R and that of larger sphere is 2R and their common surface charge density is ρ
So initial charge on smaller sphere q1=4πR2ρ
And initial charge on larger sphere Q1=16πR2ρ
Now we know that the capacitance of a sphere is proportional to its radius. So the ratio of their capacitances (smaller:larger) will be R:2R=1:2
Let the capacitance of smaller one be Cs=Cand that of larger one be Cl=2C.
After they are connected by a thin conducting wire remaining at a large distance apart their charge will be redistributed until they gain same potential.
Let this common potential be V.
By principle of conservation of charge we can write
CsV+ClV=q1+Q1=20πR2ρ
⇒CV+2CV=20πR2ρ
⇒V=20πR2ρ3C
So changed charge on large capacitor =ClV=2C×20πR2ρ3C=40πR2ρ3
So the changed surface charge density on larger capacitor
=40πR2ρ3×16πR2=56ρ
Given that the radius of smaller sphere is R and that of larger sphere is 2R and their common surface charge density is ρ
So initial charge on smaller sphere q1=4πR2ρ
And initial charge on larger sphere Q1=16πR2ρ
Now we know that the capacitance of a sphere is proportional to its radius. So the ratio of their capacitances (smaller:larger) will be R:2R=1:2
Let the capacitance of smaller one be Cs=Cand that of larger one be Cl=2C.
After they are connected by a thin conducting wire remaining at a large distance apart their charge will be redistributed until they gain same potential.
Let this common potential be V.
By principle of conservation of charge we can write
CsV+ClV=q1+Q1=20πR2ρ
⇒CV+2CV=20πR2ρ
⇒V=20πR2ρ3C
So changed charge on large capacitor =ClV=2C×20πR2ρ3C=40πR2ρ3
So the changed surface charge density on larger capacitor
=40πR2ρ3×16πR2=56ρ
