Answer
$
E_1=1.13 \times 10^4 \mathrm{~N} / \mathrm{C} .
$
Work Step by Step
The magnitude of the electric field in the gap is
$
E_1=\frac{q}{\varepsilon_0 A}=\frac{1.15 \times 10^{-9} \mathrm{C}}{\left(8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{N} \cdot \mathrm{m}^2\right)\left(115 \times 10^{-4} \mathrm{~m}^2\right)}\\=1.13 \times 10^4 \mathrm{~N} / \mathrm{C} .
$
