An empty 20 pF capacitor is charged to a potential difference of 40 V. The charging battery is then disconnected, and a piece of Teflon with a dielectric constant of 2.1 is inserted to completely fill the space between the capacitor plates.

What is the new potential difference due to insertion of the dielectric?

19 V

$\text{Charge on the Capacitor after charging is } q = CV = 800\mu C$

$\text{Charge on capacitor will remain same after inserting the dielectric}$

$ C' = kC = 42\mu C$

$ V'= \frac{q}{C'} = \frac{800 \mu C}{42\mu F}= 19 V$