Question

A conducting wire connects two charged metallic spheres A and B of radii r₁ and r₂ respectively. The distance between the spheres is very large compared to their radii. The ratio of electric fields (E_A/E_B) at the surfaces of spheres A and B will be ______.

पर्याय

• `r_1/r_2`

• `r_2/r_1`

• `r_1^2/r_2^2`

• `r_2^2/r_1^2`

Answer 1

A conducting wire connects two charged metallic spheres A and B of radii r₁ and r₂ respectively. The distance between the spheres is very large compared to their radii. The ratio of electric fields (E_A/E_B) at the surfaces of spheres A and B will be `bbunderline(r_2/r_1)`.

Explanation:

When two conducting spheres are connected by a wire, they come to the same electric potential.

V_A = V_B

For an isolated conducting sphere:

V = `(k Q)/R`

By using the equal potential condition:

`(k Q_A)/r_1 = (k Q_B)/r_2`

⇒ `Q_A/Q_B = r_1/r_2`

Electric field at the surface of a sphere:

E = `(k Q)/R^2`

So,

E_A = `(k Q_A)/r_1^2`

E_B = `(k Q_B)/r_2^2`

`E_A/E_B = Q_A/Q_B * r_2^2/r_1^2`

= `r_1/r_2 * r_2^2/r_1^2`    ...[Q_A/Q_B = r₁/r₂]

= `r_2/r_1`

∴ `E_A/E_B = r_2/r_1`