Answer:
The energy that can be stored in the capacitor is 239 nJ
Step-by-step explanation:
We first calculate the capacitance of each material. Let C₁ be the capacitance of pyrex glass and C₂ be the capacitance of polystyrene.
C₁ = κ₁ε₀A/d where κ₁ = dielectric constant of pyrex glass = 5, A = area of plates = L² where L = length of square plate = 7.40 cm = 7.40 × 10⁻² m and d = thickness of pyrex slab = 1.60 mm = 1.60 × 10⁻³ m and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m
C₁ = κ₁ε₀A/d = κ₁ε₀L²/d = 5 × 8.854 × 10⁻¹² F/m × (7.40 × 10⁻² m)²/1.60 × 10⁻³ m = 2424.2252/1.60 × 10⁻¹¹ F = 1515.14 × 10⁻¹¹ F = 15.2 × 10⁻⁹ F = 15.2 nF
C₂ = κ₂ε₀A/d where κ₂ = dielectric constant of polystyrene = 3, A = area of plates = L² where L = length of square plate = 7.40 cm = 7.40 × 10⁻² m and d = thickness of polystyrene slab = 1.60 mm = 1.60 × 10⁻³ m and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m
C₁ = κ₁ε₀A/d = κ₁ε₀L²/d = 3 × 8.854 × 10⁻¹² F/m × (7.40 × 10⁻² m)²/1.60 × 10⁻³ m = 1454.5351/1.60 × 10⁻¹¹ F = 909.08 × 10⁻¹¹ F = 9.09 × 10⁻⁹ F = 9.09 nF
Since the capacitors are in series, we find their effective capacitance C from
1/C = 1/C₁ + 1/C₂
C = C₁C₂/(C₁ + C₂)
= 15.2 × 10⁻⁹ F × 9.09 × 10⁻⁹ F/(15.2 × 10⁻⁹ F + 9.09 × 10⁻⁹ F)
= 138.168 × 10⁻¹⁸/24.29 × 10⁻⁹ F
= 5.69 × 10⁻⁹ F
The amount of electrical energy stored in a capacitor is given by W = 1/2CV² where C = capacitance and v = voltage applied. Now C = 5.69 × 10⁻⁹ F and V = 84.0 V for this capacitor
So W = 1/2 × 5.69 × 10⁻⁹ F × 84.0 V
= 238.98 × 10⁻⁹ J
≅ 239 × 10⁻⁹ J
= 239 nJ
So the energy that can be stored in the capacitor is 239 nJ