A parallel-plate capacitor features a plate area of 20 cm ^2 , a plate separation of 1.0 mm , and a dielectric slab with a dielectric constant of 5.0 filling the space between the plates. The capacitor is connected to a battery of emf 6.0 V via a 100 k resistor. Determine the energy of the capacitor 8.9 s after the connections are established.

Step-by-step solution

The capacitance of the parallel-plate capacitor with the dielectric slab is given by: C = K ₀ A d Substituting the given values: C = 5.0 8.85 10⁻¹² 20 10⁻⁴ 1.0 10⁻³ = 88.5 10⁻¹² F The time constant of the RC circuit is: = RC = (100 10^3) (88.5 10⁻¹²) = 8.85 10⁻⁶ s = 8.85 s The given time is t = 8.9 s , which is approximately equal to one time constant . The voltage across the capacitor at time t is: V = V₀ (1 - e^ -t/ ) For t : V 6.0 (1 - e⁻¹) 6.0 (1 - 0.37) = 3.78 V The energy stored in the capacitor at this time