91Ó°ÊÓ

We need to find the capacitor’s potential difference, electric field, and charge before the Mylar is inserted.

The potential difference will remain $9Volts$as the capacitor remains connected to the battery, before and after the insertion of the dielectric.

The electric field can be given as

$E=\frac{∆V}{d}=\frac{9}{0.0001}=9kV/m$,

The charge in the capacitor is given by

$q=CU=∈{∈}_0\frac{L^2}{d}U$,

where $∈$is the dielectric constant of the material. So, before insertion of the dielectric material, the charge is given as

$q=8.85·{10}^{-12}·\frac{0.{005}^2}{0.0001}·9=20pC$

We need to find the electric field, and charge after the Mylar is inserted.

The electric field will be constant even after the dielectric is inserted. It remains

$E=9kW/m$

And considering that after insertion, the Mylar the dielectric constant is $3.1$,then

$q_2=3.1·20=63pC$