A slab of a dielectric of dielectric constant 2.5 is placed in a uniform external electric field of \(5\times10^{4}\ \mathrm{V\,m^{-1}}\), with its faces normal to the field. The magnitude of the electric field produced inside the slab by its induced surface charges is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(3\times10^{4}\ \mathrm{V\,m^{-1}}\)
The induced charges on the faces of the slab set up a field opposite to the external field. In a dielectric this opposing field only reduces the external field, to \( E=E_0/K \). It does not wipe it out as it does in a conductor.
\( \begin{aligned} E &= \frac{E_0}{K} \\ &= \frac{5\times10^{4}}{2.5} \\ &= 2\times10^{4}\ \mathrm{V\,m^{-1}} \end{aligned} \)
The field of the induced charges is the part of \( E_0 \) that has been removed.
\( \begin{aligned} E_p &= E_0-E \\ &= 5\times10^{4}-2\times10^{4} \\ &= 3\times10^{4}\ \mathrm{V\,m^{-1}} \end{aligned} \)
This field points opposite to the external field.
NCERT (2026-27):
Dielectrics and Polarisation – “Unlike in a conductor, however, the opposing field so induced does not exactly cancel the external field. It only reduces it.”
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.”