A parallel-plalte capacitor has plate area `20cm^(2)`,plate separation 1.0mm and a dielectric slab of dielaectric constant 5.0 filling up the space between the plates.This capacitor is joined to a battery of emf 6.0V through a `100k(Omega)`resistor.find the energy of the capacitor `8.9(mu)s`after the connections are made.

Explanation

To solve the problem step by step, we will use the given data to find the energy stored in the capacitor after 8.9 microseconds.

### Step 1: Convert the given values to SI units
- Plate area \( A = 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2 \)
- Plate separation \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Dielectric constant \( K = 5.0 \)
- EMF of the battery \( V = 6.0 \, \text{V} \)
- Resistance \( R = 100 \, \text{k}\Omega = 100 \times 10^3 \, \Omega \)
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