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Adielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?
the electric field in the capacitor
the charge on the capacitor
the potential difference between the plates
the stored energy in the capacitor.
Solution
The charge on the capacitor
When we insert a dielectric between the plates of a capacitor, induced charges of opposite polarity appear on the face of the dielectric. They build an electric field inside the dielectric, directed
opposite to the original field of the capacitor.
Thus, the net effect is a reduced electric field.
Also, as the potential is proportional to the field, the potential decreases and so does the stored energy U, which is given by $U=\frac{q V}{2}$
Thus, only the charge on the capacitor remains unchanged, as the charge is conserved in an isolated system.