An engineer working in an electronics lab connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 275 V. Assume a plate separation of d = 1.56 cm and a plate area of A = 25.0 cm2. 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. (a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before pC after Q pC
(b) Determine the capacitance (in F) and potential difference (in V) after immersion.
(c) Determine the change in energy (in nJ) of the capacitor nJ
(d) What If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 240 V potential difference Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.)

a. i. 390.02 pC ii 31201.6 pC b. i. 1.1352 × 10⁻⁸ F  ii. 2.75 V c. -10.726 nJ

d. i. 340.54 pC ii. 27243.2 pC e. i 1.1352 × 10⁻⁸ F ii. 240 V f. 3228.319 nJ

Explanation:

a. i. charge before immersion Q

Q = CV = ε₀AV/d where ε₀ = 8.854 × 10⁻¹² F/m, A = area = 25 cm² = 25 × 10⁻⁴ m², d = plate separation = 1.56 cm = 1.56 × 10⁻² m, V = potential difference between plates = 275 V

Q = ε₀AV/d = 8.854 × 10⁻¹² F/m × 25 × 10⁻⁴ m² × 275 V/1.56 × 10⁻² m = 60871.25/1.56 × 10⁻¹⁴ C = 39020.03 × 10⁻¹⁴ C = 390.02 × 10⁻¹² C = 390.02 pC

ii. charge after immersion in water Q'

Q' = C'V = ε'ε₀AV/d = ε'Q where ε' = dielectric constant of distilled water = 80.0

Q' = ε'Q = 80 × 390.02 pC = 31201.6 pC

b. i. capacitance after immersion C'

C' = ε'ε₀A/d = 80 × 8.854 × 10⁻¹² F/m × 25 × 10⁻⁴ m²/1.56 × 10⁻² m = 17708/1.56 × 10⁻¹⁴ F = 11351.28 × 10⁻¹² F = 1.1352 × 10⁻⁸ F  

ii. The potential difference V' after immersion

Since Q' = C'V'

V' = Q'/C' = 31201.6 × 10⁻¹² C/1.1352 × 10⁻⁸ F  = 2.75 V

c. The change in energy

The initial energy of the capacitor before immersion is E = 1/2QV = 1/2 × 390.02 × 10⁻¹² C × 275 V = 107255.5 × 10⁻¹²/2 J = 53627.75 × 10⁻¹² J = 53.628 nJ

The energy of the capacitor after immersion is E' = 1/2Q'V' = 1/2 × 31201.6 ×  × 10⁻¹² C × 2.75 V = 85804.4 × 10⁻¹²/2 J = 42902.2 × 10⁻¹² J = 42.902 nJ

So ΔE = E' - E =  42.902 nJ - 53.628 nJ = -10.726 nJ

d. i with V₁ = 240 V, charge Q₁ before immersion

Q₁ = ε₀AV₁/d = 8.854 × 10⁻¹² F/m × 25 × 10⁻⁴ m² × 240 V/1.56 × 10⁻² m = 53124/1.56 × 10⁻¹⁴ C = 34053.85 × 10⁻¹⁴ C = 340.54 × 10⁻¹² C = 340.54 pC

ii charge after immersion in water Q₂ while still connected to V₁ = 240 V

Q₂ = εε₀AV₁/d = εQ₁ = 80 × 340.54 × 10⁻¹² C = 27243.2 × 10⁻¹² C = 27243.2 pC

e. i. capacitance after immersion C₁

C₁ = ε'ε₀A/d = 80 × 8.854 × 10⁻¹² F/m × 25 × 10⁻⁴ m²/1.56 × 10⁻² m = 17708/1.56 × 10⁻¹⁴ F = 11351.28 × 10⁻¹² F = 1.1352 × 10⁻⁸ F  

ii. The potential difference V after immersion

The potential difference = 240 V since the voltage is still applied after immersion.

f. The change in energy

The initial energy of the capacitor before immersion is E₁ = 1/2Q₁V₁ = 1/2 × 340.54 × 10⁻¹² C × 240 V = 81729.6 × 10⁻¹²/2 J = 40864.8 × 10⁻¹² J = 40.865 nJ

The energy of the capacitor after immersion is E₂ = 1/2Q₂V₁ = 1/2 × 27243.2 × 10⁻¹² C × 240 V = 6538368 × 10⁻¹²/2 J = 3269184 × 10⁻¹² J = 3269.184 nJ

So ΔE = E₂ - E₁ = 3269.184  nJ - 40.865 nJ = 3228.319 nJ

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