To solve this problem, we need to understand a few key concepts about charge distribution and surface charge density. Let's walk through the problem step by step.
When two conductors are brought into contact, charge will redistribute between them until they reach the same electrical potential. For spheres, the surface charge density is related to the charge and the radius of the sphere. Before contact, both spheres have the same surface charge density. Let's denote this common surface charge density as
σ .
For a sphere of radius
R, the surface area
A is given by:
A=4Ï€R2The charge
Q on each sphere can be expressed as:
Q=σ⋅A=σ⋅4πR2 Given the radii of the two spheres, the charge on each sphere before contact is:
For the sphere with radius
‌ :
Q1=σ⋅4π(‌)2=σ⋅πR2For the sphere with radius
2R :
Q2=σ⋅4π(2R)2=σ⋅16πR2 When the two spheres are brought into contact and then separated, the total charge will be redistributed between them. The total charge is:
Q‌total ‌=Q1+Q2=σ⋅πR2+σ⋅16πR2=17σ⋅πR2
Since the spheres are in contact, they will have the same potential. The potential
(V) for a sphere is given by:
V=‌ Let
Q1′ and
Q2′ be the charges on the spheres with radii
‌ and
2R respectively after they have been separated. The potential of both spheres must be equal:
‌=‌ Simplifying, we get:
And for the smaller sphere:
Q1′=‌=‌⋅‌=‌ Now, we can find the new surface charge densities:
For the sphere with radius
‌ :
σ1′=‌=‌=‌ For the sphere with radius
2R :
σ2′=‌=‌=‌=‌ Finally, the ratio of the new surface charge densities
σ1′:σ2′ is:
‌=‌=‌=4:1So, the correct answer is Option B: 4:1.