Chapter 16 - Electric Potential, Energy, and Capacitance - Learning Path Questions and Exercises - Conceptual Questions - Page 591: 20

When the power supply is not connected, the charge remains the same. But the capacitance increases once the dielectric material is inserted. Thus the potential difference decreases ($V=Q/C$) and so does the electric field ($E=V/d$). When the power supply remains connected, the potential difference remains constant, so does the electric field.

When the power supply is not connected, the charge remains the same. But the capacitance increases once the dielectric material is inserted. Thus the potential difference decreases ($V=Q/C$) and so does the electric field ($E=V/d$). When the power supply remains connected, the potential difference remains constant, so does the electric field.