College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

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

Answer

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.

Work Step by Step

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.
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