Two metal spheres, one of radius R and the other of radius 2R respectively have the same surface charge density σ. They are brought in contact and separated. What will be the new surface charge densities on them?

EXPLANATION:

Charge Q with a surface charge density σ on a spherical body of radius R is given as:

Q = σ×4πR²   -----(1)

Potential of a sphere with charge Q and radius R is given as:

V = \(\frac{Q}{4\pi \epsilon_0 R}\)   -----(2)

Given:

Radius of 1st sphere = R

Radius of 2nd sphere = 2R

Surface charge density on both 1st and 2nd sphere = σ 

Now, suppose initial charge on 1st sphere is Q1 and that on 2nd sphere is Q2. 

Using equation (1) we have,

Q1 = σ×4πR2 &  

Q2 = σ×4π(2R)2 = 4Q1  -----(3)

After separating shperes when brought in contact both have same potential.

Let us suppose final charges on two spheres are Q'₁ & Q'2 .

∴ V₁ = V₂  ⇒ \(\frac{Q'_1}{4\pi \epsilon_0 R}\) = \(\frac{Q'_2}{4\pi \epsilon_0 (2R)}\)    (using equation (2))

⇒Q'2 = 2Q'1   -----(4)

∵ Total charge on the sphere will be same before and after separation.

∴ Q1 + Q2 = Q'1 + Q'2

⇒Q1 + 4Q1 = Q'1 + 2Q'1

⇒5Q1 = 3Q'1   ⇒Q'1= \(\frac{5}{3}\)Q1 = \(\frac{5}{3}\)×σ×4πR2

or, σ₁ = \(\frac{Q'_1}{4\pi \epsilon_0 R^2}\) = \(\frac{5}{3}\)σ  

Simillarly, σ2 =  \(\frac{5}{6}\)σ 

Hence option 2) is the correct choice.