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

The corrrect answer is Option 3: 19 V

Given:
C = 20 pF = 20 × 10⁻¹² F
V = 40 V
k = 2.1

Charge on capacitor:
q = CV = (20 × 10⁻¹²)(40) = 800 × 10⁻¹² C

After inserting dielectric:
C′ = kC = 2.1 × 20 × 10⁻¹² = 42 × 10⁻¹² F

Since charge remains constant:
V′ = q / C′ = (800 × 10⁻¹²) / (42 × 10⁻¹²) = 800 / 42 ≈ 19 V

Therefore, V′ ≈ 19 V