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

Answer:

the answer will be 2.9 vs

Answer 2

After the connections are made, the energy of the capacitor is 6.3$\times$10-10 J.

Explanation:

Given:

Plates’ area, A = 20 cm2

Between the plates separation, d = 1 mm

Dielectric constant, k = 5

Battery’s Emf, E = 6 V

Circuit’s resistance, R = 100 $\times$ 103 Ω

The parallel-plate capacitor’s capacitance,

C = K∈0Ad = 10 $\times$ 8.85 $\times$ 10-12 $\times$ 20 $\times$ 10-41 $\times$ 10-3 = 88.5 $\times$ 10-12 C

The growth of charge through the capacitor after the connections are made,

Q = EC (1 − e-tRC)

= 6 $\times$ 88.5 $\times$ 10−12 (1 − e-8.98.85)

= 335.6 $\times$ 10−12 C

So, in the capacitor, the energy stored,

U = 12Q2C = 6.3$\times$10-10 J.