The plates of a charged parallel plate capacitor have surface charge densities \(\pm8\times10^{-6}\ \mathrm{C\,m^{-2}}\). A dielectric of dielectric constant 4 now fills the space between the plates completely. The magnitude of the induced surface charge density on each face of the dielectric is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(6\times10^{-6}\ \mathrm{C\,m^{-2}}\)
The field polarises the dielectric. Its two faces then carry induced charge densities \( -\sigma_p \) and \( +\sigma_p \), opposite in sign to the plates next to them. So the net charge density that sets the field is \( \sigma-\sigma_p \), and this equals \( \sigma/K \).
\( \begin{aligned} \sigma-\sigma_p &= \frac{\sigma}{K} \\ \sigma_p &= \sigma\left(1-\frac{1}{K}\right) \\ &= 8\times10^{-6}\times\left(1-\frac{1}{4}\right) \\ &= 6\times10^{-6}\ \mathrm{C\,m^{-2}} \end{aligned} \)
From the above, \( \sigma_p=6\times10^{-6}\ \mathrm{C\,m^{-2}} \). Hence, the induced charge on each face of the dielectric is three-fourths of the free charge on the plate next to it.
NCERT (2026-27):
Effect of Dielectric on Capacitance – “… The electric field in the dielectric then corresponds to the case when the net surface charge density on the plates is ±(σ – σp). …”
Effect of Dielectric on Capacitance – “For linear dielectrics, we expect σp to be proportional to E0, i.e., to σ. Thus, (σ – σp) is proportional to σ and we can write”
\( \displaystyle \sigma-\sigma_P=\frac{\sigma}{K} \)\( \qquad (2.49) \)