Question

Two parallel-plate capacitors C₁ and C₂ are connected in series to a battery. Both capacitors have the same plate area of 3.40 cm² and plate separation of 2.65 mm. However, the first capacitor C₁ is filled with air, while the second capacitor C₂ is filled with a dielectric that has a dielectric constant of 3.40. The total charge on the series arrangement is 13.8 pC.

(a) What is the battery voltage?
V

(b) What is the potential difference across each capacitor?

ΔV₁ =   V

ΔV₂ =   V

Answer 1

Since the capacitors are in series, the total charge on each capacitor is same.
Consider Q be the charge on each capacitor, 2Q = 13.8 pC
Q = 6.9 pC
Capacitance of a capacitor can be written as C = KoA/d
Where K is the dielectric constant, o is the vacuum permittivity, A is the area of the metal plates and d is the separation between the metal plates.
A = 3.4 x 10^-4 m².
d = 2.65 x 10^-3 m
C1 = [1 x (8.85 x 10^-12) x (3.4 x 10^-4)] / (2.65 x 10^-3)
= 1.1355 pF
Similarly,
C2 = [3.4 x (8.85 x 10^-12) x (3.4 x 10^-4)] / (2.65 x 10^-3)
= 3.86 pF

V1 = Q/C1
= (6.9 pC) / (1.1355 pF)
= 6.08 V
V2 = Q/C2
= (6.9 pC) / (3.86 pF)
= 1.79 V

Voltage of the battery = 1 + V2
= 7.86 V