Q 12-02-043JEE MainJEE Main 2026 (2 Apr, Shift 1)Easy
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 $E_{S1}$ and $E_{S2}$ respectively. The value of $\dfrac{E_{S1}}{E_{S2}}$ is ______.
Answer: (D) $\dfrac94$
Connected conductors come to the same potential: $V=\dfrac{kq_1}{r_1}=\dfrac{kq_2}{r_2}$.
Field at the surface: $E=\dfrac{kq}{r^2}=\dfrac{V}{r}$, so $E\propto\dfrac1r$.
$$\frac{E_{S1}}{E_{S2}}=\frac{r_2}{r_1}=\frac{18}{8}=\frac94$$
Solution by Sreeraj P, M.Sc Physics