Case: Read the passage and answer the question.
In a parallel plate air capacitor having plate separation 5 cm, an electric field of \( 4\times10^{4}\,\mathrm{V/m} \) is established between the plates. After the removal of the battery, a metal plate of thickness t = 2 cm is inserted between the plates of the capacitor.
What is the potential difference across the capacitor if a dielectric slab of dielectric constant K = 2 and the same thickness were inserted in place of the metal plate?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1600 V
After the battery is removed, the charge on the plates stays constant.
So the field in the air gap stays \( E_0=\sigma/\varepsilon_0 \).
Inside a dielectric slab the field is reduced to \( E_0/K \).
The potential difference is the sum of field × distance over the air part and the slab.
In SI units the data are:
\( E_0=4\times10^{4}\,\mathrm{V/m} \)
\( K=2 \)
\( d=5\,\mathrm{cm}=0.05\,\mathrm{m} \)
\( t=2\,\mathrm{cm}=0.02\,\mathrm{m} \)
The potential difference across the air gaps is:
\( \begin{aligned} V_{\mathrm{air}} &= E_0\,(d-t) \\ &= 4\times10^{4}\,\mathrm{V/m}\times0.03\,\mathrm{m} \\ &= 1200\,\mathrm{V} \end{aligned} \)
The potential difference across the slab is:
\( \begin{aligned} V_{\mathrm{slab}} &= \frac{E_0}{K}\,t \\ &= \frac{4\times10^{4}\,\mathrm{V/m}}{2}\times0.02\,\mathrm{m} \\ &= 400\,\mathrm{V} \end{aligned} \)
Adding the two gives the total:
\( \begin{aligned} V &= V_{\mathrm{air}}+V_{\mathrm{slab}} \\ &= 1200\,\mathrm{V}+400\,\mathrm{V} \\ &= 1600\,\mathrm{V} \end{aligned} \)