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,

\[C = \frac{K \in_0 A}{d}\]

\[ = \frac{5 \times 8 . 85 \times {10}^{- 12} \times 20 \times {10}^{- 4}}{1 \times {10}^{- 3}}\]

\[ = \frac{10 \times 8 . 85 \times {10}^{- 12} \times 20 \times {10}^{- 4}}{1 \times {10}^{- 3}}\]

\[ = 88 . 5 \times {10}^{- 12} C\]

After the connections are made, growth of charge through the capacitor,

\[Q = EC\left(1 − e^{- \frac{t}{RC}}\right)\]

\[=6\times88.5\times10^{-12}\left(1-e^{- \frac{8 . 9}{8 . 85}}\right)\]

= 335.6 × 10⁻¹² C

Thus, energy stored in the capacitor,

\[U = \frac{1}{2}\frac{Q^2}{C}\]

\[ = \frac{1}{2} \times \frac{335 . 6 \times 335 . 6 \times {10}^{- 24}}{88 . 5 \times {10}^{- 12}}\]

\[ = \frac{335 . 6 \times 335 . 6}{88 . 5 \times 2} \times {10}^{- 12} \]

\[ = 6 . 3 \times {10}^{- 10} J\]