Question

A parallel-plate capacitor has plate area 20 cm², plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy of the capacitor 8.9 μs after the connections are made.

Answer 1

Given:
Area of the plates, A = 20 cm²
Separation between the plates, d = 1 mm
Dielectric constant, k = 5
Emf of the battery, E = 6 V
Resistance of the circuit, R = 100 × 10³ Ω
The capacitance of a parallel-plate capacitor,
$$\begin{gathered} C=\frac{K{\mathit{\in }}_{0}A}{d} \\ =\frac{5\times 8.85\times {10}^{-12}\times 20\times {10}^{-4}}{1\times {10}^{-3}} \\ =\frac{10\times 8.85\times {10}^{-12}\times 20\times {10}^{-4}}{1\times {10}^{-3}} \\ =88.5\times {10}^{-12}\mathrm{C} \end{gathered}$$
After the connections are made, growth of charge through the capacitor,
Q = EC (1 − ${\mathrm{e}}^{-\frac{t}{RC}}$)
= 6 × 88.5 × 10⁻¹² (1 − $e^{-\frac{8.9}{8.85}}$)
= 335.6 × 10⁻¹² C
Thus, energy stored in the capacitor,
$$\begin{gathered} U=\frac{1}{2}\frac{Q^2}{C} \\ =\frac{1}{2}\times \frac{335.6\times 335.6\times {10}^{-24}}{88.5\times {10}^{-12}} \\ =\frac{335.6\times 335.6}{88.5\times 2}\times {10}^{-12} \\ =6.3\times {10}^{-10}\mathrm{J} \end{gathered}$$