Question

Two insulated charged spherical bodies having radii R and 2R have their same surface density of charges (D). When these two bodies are connected by a conducting wire, what will be the surface density of charge of the bigger sphere?

Answer 1

`5/6 D`

Explanation:

Initial amount of charge will be conserved:

`Q = q_1 + q_2 = 4 pi R^2 D + 4 pi (2R)^2 D = 20 pi R^2 D`

`= q_1' + q_2' = Q'`

Current will flow until the potential on the surface of each sphere is the same.

Potential on surface of sphere is: `V = (k q)/r`

• `V_1' = V_2' implies (q_1')/(r_1) = (q_2') /(r_2)`

• `(q_1')/R = (q_2') /(2R) implies q_1' = (q_2')/2`

Therefore:

`(3q_2')/2 = 20 pi R^2 D implies qquad [(q_1' = 20/3 pi R^2 D),(q_2' = 40/3 pi R^2 D) :}`

`D' = (Q)/A = Q/(4 pi r^2)`

`implies qquad [(D_1' = 5/3 D),(color(red)(D_2' = 5/6 D)) :}`