Explanation

(a) In free space, the electric field between plates of capacitor.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{0} = \frac{q}{A\varepsilon_{0}}\ \ \ $ ...(i)
When plates of capacitor dipped in oil tank then, the electric field between the plates is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{0} = \frac{q}{A\varepsilon}\ $ …(ii)
(when $\varepsilon$ is the permittivity of medium)
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E\ = \frac{E_{0}}{AK\varepsilon_{0}}$ $\lbrack\therefore\ \varepsilon = K\varepsilon_{0}\rbrack$
From Eqs. (i) and (ii),
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \frac{E_{0}}{K}$
(where $K$ is the dielectric constant )
Hence, the electric field between the plates is increase.