Conductors And Dielectrics

When a dielectric of relative permittivity $\kappa$ is inserted into a charged, isolated capacitor, the bound charges in the dielectric polarise and set up an opposing field, reducing the net field to $E=\frac{E_0}{\kappa }$, as explained in NCERT Class 12, Chapter 2 (Electrostatic Potential and Capacitance). Since the capacitor is disconnected, the free charge on the plates stays constant, and with $\kappa =5$ the field becomes one fifth of the vacuum value $E_0$. The option that the field increases five times is wrong because polarisation always weakens rather than strengthens the field inside a dielectric. The option that the field is unchanged is wrong because that would ignore the polarisation charges entirely. The option that the field drops to zero is wrong because a dielectric only partially cancels the field, unlike a conductor which cancels it completely. Because the free charge on the isolated plates is unchanged, the reduced field also means the voltage across the plates drops and the capacitance correspondingly rises by the same factor. A magnitude and physical check confirms the reduction: the induced surface charges on the dielectric oppose the plate charges but can never exceed them, so the net field decreases by the factor $\kappa$ without ever reaching zero.