Two metal spheres, one fo radius R and the other of radius 2R, both have same surface charge density s. They are brought in contact and seprated. What will be new surface charge densitites on them ?

Explanation

Radius of sphere =R
Surface charge density on sphere `A= sigma`
Radius of sphere B = 2R
Surface charge density on sphere `B= sigma`
Before contact, the charge on sphere A is`Q_1` = Surface charge density × Surface area ... (i)
Before contact, the charge on sphere B is `Q_2` = Surface charge density × Surface area
Let after the contact, the charge on A is `Q_1` and the charge on B is `Q_2` . According to the conservation of charge, the charge before contact is equal to charge after contact.
`Q_1+Q_2=Q_1+Q_2`Putting the values of `Q_1 and Q_2`from Eqs. (i) and (ii), we get
`Q_1+Q_2=4piR^2 sigma+16piR^2 sigma=20 pi R^2 sigma`..........(iii)
As they are in contact. So, they have same potential.
Potential on sphere A is `V_A=1/(4pie_0).Q_1/R` and Potential on sphere B is `V_B=1/(4pi e_0).Q_2/(2R)`
So `V_A=V_B 1/(4pi e_0). Q_1/R=1/(4 pi e_0).Q_2/(2R)`
`Q_1/R=Q_2/(2R) 2Q_1=Q_2`
Putting the value of `Q_2` in Eq (iii) we get `Q_1+2Q_1=20 pi R^2 sigma`
`3Q_1=20 pi R^2 sigma. Q_1=20/3 pi R^2 sigma and Q_2 =40/3 pi R^2 sigma`
Let the new charge densities be `sigma_1 and sigma_2`
`simga_1=Q_1/(4pi R^2)=(20 pi R^2 sigma)/(3 times 4 pi R^2)=5/3 sigma`
`sigma_2=Q_2/(4pi(2R)^2)=(40 pi R^2 sigma)/(3 times 4pi times 4R^2)=(40 sigma)/(16 times 3)`
`sigma_2=(10 sigma)/(4 times 3)=5/6 sigma`
Thus, the surface charge densities on spheres after contacting are `5/3 sigma and 5/6 sigma`