Case : Read the passage and answer the question(s).

In a parallel plate air capacitor having plate separation 0.05mm, an electric field of 4X10⁴ V/m is established between the plates. After the removal of the battery a metal plate of thickness t = 0.02 mm is inserted between the plates of the capacitor. 

What is the potential difference across capacitor before the introduction of metal plates?

2 V

The correct answer is Option 4: 2 V

V = E x d 

Electric Field ($E$): $4 \times 10^4 \text{ V/m}$

Plate Separation ($d$): $0.05 \text{ mm}$ 

$d = 0.05 \text{ mm} = 0.05 \times 10^{-3} \text{ m}$ (or $5 \times 10^{-5} \text{ m}$)