An isolated parallel plate air capacitor has its plates 2.5 mm apart and a potential difference of 200 V across it. A dielectric slab is then placed so that it completely fills the space between the plates. The electric field inside the slab is found to be \(2\times10^{4}\ \mathrm{V\,m^{-1}}\). The dielectric constant of the slab is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(4\)
With air between the plates, the field is the potential difference divided by the gap. The capacitor is isolated, so the charge stays the same when the slab goes in. The field inside the slab is then \( E_0/K \).
\( \begin{aligned} E_0 &= \frac{V_0}{d} \\ &= \frac{200}{2.5\times10^{-3}} \\ &= 8\times10^{4}\ \mathrm{V\,m^{-1}} \end{aligned} \)
\( \begin{aligned} K &= \frac{E_0}{E} \\ &= \frac{8\times10^{4}}{2\times10^{4}} \\ &= 4 \end{aligned} \)
NCERT (2026-27):
Example 2.8 – “Let E0 = V0/d be the electric field between the plates when there is no dielectric and the potential difference is V0. If the dielectric is now inserted, the electric field in the dielectric will be E = E0/K.”