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 capacitance of the condenser?

20 pF 42 pF 40 pF 100 pF

42 pF

$C'= kC = 2.1 \times 20pF = 42pF$