The plates of an isolated parallel plate capacitor carry surface charge densities \(\pm8.85\times10^{-7}\ \mathrm{C\,m^{-2}}\). The space between the plates is completely filled with a dielectric of dielectric constant 5. The electric field inside the dielectric is (take \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2\times10^{4}\ \mathrm{V\,m^{-1}}\)
With vacuum between the plates, the field is \( E_0=\sigma/\varepsilon_0 \). The polarised dielectric sets up a field that opposes \( E_0 \). So the field inside the dielectric falls to \( E_0/K \).
\( \begin{aligned} E_0 &= \frac{\sigma}{\varepsilon_0} \\ &= \frac{8.85\times10^{-7}}{8.85\times10^{-12}} \\ &= 1\times10^{5}\ \mathrm{V\,m^{-1}} \end{aligned} \)
\( \begin{aligned} E &= \frac{E_0}{K} \\ &= \frac{1\times10^{5}}{5} \\ &= 2\times10^{4}\ \mathrm{V\,m^{-1}} \end{aligned} \)
NCERT (2026-27):
Effect of Dielectric on Capacitance – “… When there is vacuum between the plates,”
\( \displaystyle E_0=\frac{\sigma}{\varepsilon_0} \)
Example 2.8 – “… If the dielectric is now inserted, the electric field in the dielectric will be E = E0/K.”