Question

Two spherical conductors A and B of radii $1$ mm and $2$ mm 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. $1:4$

Answer 1

The correct option is C $2:1$

$\text{Step 1: Charges on spheres after connecting with wire[Refer Fig.]}$

After connection, at steady state final potential of spheres will be same.

Therefore, $V_1=V_2$
$\Rightarrow K\frac{Q_1}{r_1}=K\frac{Q_2}{r_2}$

$\Rightarrow \frac{Q_1}{r_1}=\frac{Q_2}{r_2}$

$\Rightarrow \frac{Q_1}{Q_2}=\frac{r_1}{r_2}$ $....\left(1\right)$

$\text{Step 2: Ratio of electric fields}$

The ratio of electric fields:

$\frac{E_1}{E_2}=\frac{K\frac{Q_1}{r_1^2}}{K\frac{Q_2}{r_2^2}}$

$\Rightarrow \frac{E_1}{E_2}=\frac{Q_1}{r_1^2}\times \frac{r_2^2}{Q_2}$

From equation $\left(1\right)$

$\Rightarrow \frac{E_1}{E_2}=\frac{r_1\times r_2^2}{r_1^2\times r_2}\Rightarrow \frac{E_1}{E_2}=\frac{r_2}{r_1}=\frac{2}{1}$ (Given $r_1=1mm$, $r_2=2mm$)

Hence, option C is correct.

$\text{Note:}$ Since the distance between the spheres is large as compared to their diameters, the induced effects may be ignored.