Take $\mathrm{O}$ to be the centre of the circle of radius $\mathrm{R}$ and angle $\mathrm{CAB}=60^{\circ}$
Figure:$Image$ 
(Assume, $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^{+9}\; Nm ^2 C ^{-2}$)

Reason : The field just outside the capacitor is $\frac{\sigma }{{{\varepsilon _0}}}$. ( $\sigma $ is the charge density).

