The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \mathrm{\AA }$ apart is,
$\left(m_{\mathrm{e}}\simeq 9\times {10}^{-31} \mathrm{kg}, e=1.6\times {10}^{-19} \mathrm{C}\right)$ (Take $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9 \mathrm{N} {\mathrm{m}}^2 {\mathrm{C}}^{-2}$)

A parallel plate capacitor having cross-sectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of the same area but the thickness, $\frac{d}{2}$, is inserted between the plates as shown in figure having dielectric constant $K(=4).$ The ratio of new capacitance to its original capacitance will be,