Question

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

Answer 1

Let the two metal spheres have radii is R and 2R.

Initial surface charge density on both spheres is σ.

We have to find the new surface charge densities after they are brought into contact and then separated.

Surface charge density is given by:

σ = `Q/(4pir^2)`

Q = σ(4πr²)

Charge on sphere of radius R:

Q₁ = σ(4πR²)

Charge on sphere of radius 2R:

Q₂ = σ(4πR²)

= σ[4π(2R)²]

= σ(4π4R²)

= σ(16πR²)

Q₂= 4Q₁

`Q_"Total" = Q_1 + Q_2`

= σ(4πR²) + σ(16πR²)

= 20πσR²

When conducting spheres are brought in contact, their potentials become equal.

Potential of a conducting sphere:

`V = 1/(4piε_0)Q/R`

`(Q"'"_1)/R = (Q"'"_2)/R`

Q'2 = 2Q'1

Apply conservation of charge:

Q'1 + Q'2 = 20πσR² 

Substitute Q'2 = 2Q'1

Q'1 + 2Q'1 = 20πσR² 

3Q'1 = 20πσR² 

Q'1 = `20/3πσR^2`

and Q'2 = 2Q'1

= `2 xx 20/3πσR^2`

= `40/3πσR^2`

New surface charge densities for a sphere of radius R,

`σ_1 = (Q"'"_1)/(4piR^2)`

`σ_1 = (20/3πσR^2)/(4piR^2)`

`σ_1 = (20πσR^2)/(3 xx 4piR^2)`

`σ_1 = 5/3 σ`

For a sphere of radius 2R,

`σ_2 = (Q"'"_2)/(4pi(2R)^2)`

`σ_2 = (40/3πσR^2)/(4pi4R^2)`

`σ_2 = (40/3πσR^2)/(16piR^2)`

`σ_2 = (40πσR^2)/(3 xx 16piR^2)`

`σ_2 = 5/6 σ`