Two charged conducting spheres $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ 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 $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ spheres are $\mathrm{E}_{\mathrm{S} 1}$ and $\mathrm{E}_{\mathrm{S} 2}$ respectively. The value of $\frac{\mathrm{E}_{\mathrm{S} 1}}{\mathrm{E}_{\mathrm{S} 2}}$ is ____.

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Electrostatics

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 ____.

Answer: (D)

Connected by a wire, the spheres reach a common potential,

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

, so

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

. The surface field of a sphere is

E = \frac{k Q}{r ^{2}}

, giving

\frac{E _{S 1}}{E _{S 2}} = \frac{Q _{1}}{Q _{2}} (\frac{r _{2}}{r _{1}})^{2} = \frac{r _{1}}{r _{2}} (\frac{r _{2}}{r _{1}})^{2} = \frac{r _{2}}{r _{1}} = \frac{18}{8} = \frac{9}{4}

. The ratio

E_{S 1} / E_{S 2}

\frac{9}{4}

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Single correct option