The plates of a parallel plate capacitor have an area of `90 cm^(2)` each and are separated by 2.5mm. The capacitance is charged by connecting it to a 400V supply.
(a) How much electrostatic energy is stored by the capacitor ?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume (u). Hence arrive at a relation between U and the magnitude of electric field E between the plates.

Explanation

To solve the given problem step by step, we will break it down into parts (a) and (b) as specified in the question.

### Part (a): Calculate the Electrostatic Energy Stored by the Capacitor

1. **Identify the given values:**
- Area of the plates, \( A = 90 \, \text{cm}^2 = 90 \times 10^{-4} \, \text{m}^2 = 0.009 \, \text{m}^2 \)
- Separation between the plates, \( d = 2.5 \, \text{mm} = 2.5 \times 10^{-3} \, \text{m} \)
- Voltage, \( V = 400 \, \text{V} \)
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