Charged spheres

The outside electric field of a sphere follows frow Gauss' law $\begin{aligned} \oint \vec E \cdot \vec A &= E \cdot 4 \pi r^2 = \frac{Q}{\varepsilon_0} \\ \Rightarrow \quad E &= \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \end{aligned}$ since the field $\vec E(\vec r) = E(r) \vec r$ has spherical symmetry. Therefore, the electric potential equals $\phi(\vec r) = \int_{\vec r}^\infty \vec E \cdot d \vec r' = \int_{r}^\infty E(r') d r' = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$ Since $\phi_1 = \phi_2$ for both spheres it follows $\begin{aligned} \frac{Q_1}{Q_2} &= \frac{r_1}{r_2} & \text{or} \quad Q_i &\propto r_i \\ \frac{E_1}{E_2} &= \frac{r_2}{r_1} & \text{or} \quad E_i &\propto \frac{1}{r_i} \\ \frac{\sigma_1}{\sigma_2} &= \frac{r_2}{r_1} & \text{or} \quad \sigma_i &\propto \frac{1}{r_i} \\ \end{aligned}$ Therefore, $Q_1 > Q_2, \quad E_1 < E_2 \quad \text{and} \quad \sigma_1 < \sigma_2$