Two charge metallic spheres are joined by a very thin metal wire. If the radius of the larger sphere is four times that of the smaller sphere, the electric field near the larger sphere is

Since the spheres are joined by a metal wire, their potentials will be same.
Let the radius of the smaller sphere be \(r_1\) and the radius of the larger sphere be \(r_2=4 r_1\) the \(\mathrm{v}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1}{\mathrm{r}_1}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_2}{\mathrm{r}_2}\)
\(
\therefore \frac{\mathrm{q}_2}{\mathrm{q}_1}=\frac{\mathrm{r}_2}{\mathrm{r}_1}
\)
The electric field near the surface of spheres are given by
\(
\begin{aligned}
& \mathrm{E}_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1}{\mathrm{r}_1^2} \text { and } \mathrm{E}_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_2}{\mathrm{r}_2^2} \\
& \therefore \frac{\mathrm{E}_2}{\mathrm{E}_1}=\frac{\mathrm{q}_2}{\mathrm{q}_1} \cdot \frac{\mathrm{r}_1^2}{\mathrm{r}_2^2}=\frac{\mathrm{r}_2}{\mathrm{r}_1} \cdot \frac{\mathrm{r}_1^2}{\mathrm{r}_2^2}=\frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{1}{4}
\end{aligned}
\)