Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is :

Question

Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is : (A) 4: 1 (B) 1: 2 (C) 2: 1 (D) 1: 4

Tardigrade Admin · Accepted Answer

When the two conducting spheres are connected by a conducting wire, charge will flow from one sphere (having higher potential) to other (having lower potential) till both acquire the same potential.
After connection,

V_{1} = V_{2}

\Rightarrow K \frac{Q _{1}}{r _{1}} = K \frac{Q _{2}}{r _{2}}

\Rightarrow \frac{Q _{1}}{r _{1}} = \frac{Q _{2}}{r _{2}}

The ratio of electric fields

\frac{E _{1}}{E _{2}} = \frac{K \frac{Q _{1}}{r _{1}^{2}}}{K \frac{Q _{2}}{r _{2}^{2}}}

\Rightarrow \frac{E _{1}}{E _{2}} = \frac{Q _{1}}{r _{1}^{2}} \times \frac{r _{2}^{2}}{Q _{2}}

\Rightarrow \frac{E _{1}}{E _{2}} = \frac{r _{1} \times r _{2}^{2}}{r _{1}^{2} \times r _{2}} \Rightarrow \frac{E _{1}}{E _{2}} = \frac{r _{2}}{r _{1}} = \frac{2}{1}

$4 : 1$

$1 : 2$

$2 : 1$

$1 : 4$

Q. Two spherical conductors A and B of radii 1mm and 2mm are separated by a distance of 5cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is :

4:1

1:2

2:1

1:4

$4 : 1$

$1 : 2$

$2 : 1$

$1 : 4$