Case: Read the passage and answer the question.
In a parallel plate air capacitor having a plate separation of 0.05 mm, an electric field of \( 4\times10^{4}\,\mathrm{V/m} \) is established between the plates by a battery. After the battery is removed, a metal plate of thickness t = 0.02 mm is inserted between the plates of the capacitor, parallel to them.
What is the potential difference across the capacitor before the introduction of the metal plate?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2.0 V
Between the plates of a parallel plate capacitor the electric field is uniform. So the potential difference is the field times the plate separation, \( V=Ed \). Here \( d=0.05\,\mathrm{mm}=5\times10^{-5}\,\mathrm{m} \).
\( \begin{aligned} V &= Ed \\ &= 4\times10^{4}\,\mathrm{V/m}\times5\times10^{-5}\,\mathrm{m} \\ &= 2.0\,\mathrm{V} \end{aligned} \)