A student working in the physics laboratory connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 270 V. Assume a plate separation of d = 1.40 cm and a plate area of A = 25.0 cm^2. When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0 Required:
a. Calculate the charge on the plates (in pC) before and after the capacitor is submerged.
b. Determine the capacitance (in F) and potential difference (in V) after immersion
c. Determine the change in energy (in nJ) of the capacitor

a)  Q₀ = Q = 4.27 10² pC,  b) C = 1.26 10⁻¹⁰ F,  ΔV = 3.375 V, c) ΔE = 56.88  nJ

Explanation:

a) the capacitance of a parallel plate capacitor is

        C = ε₀ A / d

let's reduce the magnitudes to the SI system

        d = 1.40 cm = 0.0140 m

        A = 25.0 cm² (1 m / 100 cm) ² = 25.0 10⁻⁴ m²

         C = 8.85 10⁻¹² 25.0 10⁻⁴ / 0.0140

         C = 1.58 10⁻¹² F

the capacitance is also

          C = Q / ΔV

          Q = V ΔV

          Q = 270  1.58 10⁻¹²

          Q = 4.27 10⁻¹⁰ C

When the battery is removed and the capacitor is inserted into the dielectric, the charge should remain the same,

           Q₀ = Q = 4.27 10⁻¹⁰ C

let's reduce to pC

           Q₀ = 4.27 10⁻¹⁰ C (10¹² pC / 1C)

           Q₀ = Q = 4.27 10² pC

b) After being immersed in the fluid of constant k = 80.0

capacitance is

            C = k Co

            C = 80 1.58 10⁻¹²

            C = 1.26 10⁻¹⁰ F

the voltage difference is

            ΔV = ΔV₀ / k

            ΔV = 270/80

            ΔV = 3.375 V

c) We stop the energy before the dive

             E₀ = ½ C₀ ΔV₀²

after the dive

              E = ½ (k C₀) (ΔV₀/k)²

              E = ½ C₀ ΔV₀² / k

              E = E₀ / k

the change in energy is

            ΔE = E -E₀

            ΔE = E₀ / k - E₀

            ΔE = E₀ ( )

we calculate

             E₀ = ½ 1.58 10⁻¹² 270²

             E₀ = 5.76 10⁻⁸ J

             ΔE = 5.76 10⁻⁸ ()

             ΔE = 5.688 10⁻⁸ J

we reduce to nJ

             ΔE = 5.688 10⁻⁸ J (10⁹ nJ / 1J)

             ΔE = 56.88  nJ

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge