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?
Answer & explanation
Correct answer: option 4
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 $5 \times 10^{-5} \text{ m}$)
$V = (4 \times 10^4 \text{ V/m}) \times (5 \times 10^{-5} \text{ m})$
$V = 20 \times 10^{-1} \text{ V}$
$V = 2 \text{ V}$