Reason : The field just outside the capacitor is $\frac{\sigma }{{{\varepsilon _0}}}$. ( $\sigma $ is the charge density).
Draw a Gaussian surface $A B C D$ as shown.
The field $\overrightarrow{\mathrm{E}}$ is uniform on faces $\mathrm{AD}$ and $B C$
$ \Rightarrow \,\oint {\overrightarrow {\text{E}} \cdot {\text{ds}} = 0} $ yields $\vec {\mathrm{E}}=0$


