JEE Main 2026 — Apr 2, Shift 1 — Physics solution

JEE Main2026Apr 2, Shift 1Physics

Two charged conducting spheres $S_{1}$ and $S_{2}$ of radii $8$ cm and $18$ cm are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on $S_{1}$ and $S_{2}$ spheres are $E_{S_{1}}$ and $E_{S_{2}}$ respectively. The value of $\frac{E _{S_{1}}}{E _{S_{2}}}$ is ________.

A
$\frac{3}{2}$
B
$\frac{2}{3}$
C
$\frac{4}{9}$
D
$\frac{9}{4}$

Solution

Assuming the spheres are placed far apart, connecting them by a wire equalises their potentials:

$\frac{k Q _{1}}{r _{1}} = \frac{k Q _{2}}{r _{2}} \Rightarrow \frac{Q _{1}}{Q _{2}} = \frac{r _{1}}{r _{2}} = \frac{8}{18} = \frac{4}{9}$

The surface field of an isolated sphere is $E = \frac{k Q}{r ^{2}}$ . Hence:

$\frac{E _{S_{1}}}{E _{S_{2}}} = \frac{Q _{1}}{Q _{2}} \cdot (\frac{r _{2}}{r _{1}})^{2} = \frac{4}{9} \times (\frac{9}{4})^{2} = \frac{9}{4}$

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