Two parallel-plate capacitors C1 and C2 are connected in series to a battery. Both capacitors have the same plate area of 3.20 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 2.30. 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

Please show all the work done

Explanation:

capacitance of air capacitor = ε A / d , where A is plate area , d is plate separation , ε = 8.85 x 10⁻¹² .

replacing the given data

capacitance = (8.85 x 10⁻¹²  x 3.2 x 10⁻⁴ ) / 2.65 x 10⁻³

C₁ = 10.68 x 10⁻¹³ F

capacitance of capacitor with dielectric = 2.3 x 10.68 x 10⁻¹³

C₂ = 24.58 x 10⁻¹³  F .

Total charge = 13.8 x 10⁻¹² C

Total capacitance = 10.68 x 10⁻¹³ +  24.58 x 10⁻¹³

= 35.26 x 10⁻¹³ F

battery voltage =  total charge / total capacity

= 13.8 x 10⁻¹² / 35.26 x 10⁻¹³

= 3.9  V

b ) Since both the capacitance are joined to battery in parallel , both will acquire the potential of the battery . Hence potential difference across each capacitor will be 3.9 V .