Question

A metallic sphere A isolated from ground is charged to +50 μC. This sphere is brought in contact with other isolated metallics sphere B of half the radius of sphere A. The charge on the two-sphere will be now in the ratio

विकल्प

• 1 : 2

• 2 : 1

• 4 : 1

• 1 : 1

Answer 1

2 : 1

Explanation:

Two metallic spheres brought together in contact with each other.

Radius of A = r₁

Radius of B = r₂

also r₁ = 2 r₂

A charge will flow from sphere A to sphere B till the potential becomes the same for both A and B.

∴ V_A = V_B

`∴ 1/(4piepsilon_0) "q"_"A"/"r"_1 = 1/(4piepsilon_0) "q"_"B"/"r"_2`

`∴ cancel(1/(4piepsilon_0)) "q"_"A"/"r"_1 = cancel(1/(4piepsilon_0)) "q"_"B"/"r"_2`

`∴ "q"_"A"/r_1 = q_"B"/r_2`

`∴ "q"_"A" = r_1/r_2 "q"_"B"`

`∴ "q"_"A" = (2 r_2)/r_2"q"_"B"`

`∴ "q"_"A" = 2" q"_"B"`

`∴ "q"_"A"/ " q"_"B" = 2/1 ~~ 2 : 1`