A parallel-plate capacitor has square plates that are 7.40 cm on each side and 3.20 mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 7.40 cm on a side and 1.60 mm thick. One slab is Pyrex glass and the other slab is polystyrene. If the potential difference between the plates is 84.0 V, find how much electrical energy (in nJ) can be stored in this capacitor.

The energy that can be stored in the capacitor is 239 nJ

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

waste

explanation: