Question

Two identical air filled parallel plate capacitors are charged to the same potential in the manner shown by closing the switch S. If now the switch S is opened and the space between the plates is filled with a dielectric of relative permittivity ${\epsilon }_r$, then :

• A. The potential difference as well as charge on each capacitor goes up by a factor ${\epsilon }_r$
• B. The potential difference as well as charge on each capacitor goes down by a factor ${\epsilon }_r$
• C. The potential difference across A remains constant and the charge on B remains unchanged
• D. The potential difference across B remains constant while the charge on A remains unchanged

Answer 1

Answer: (C)

The potential difference across A remains constant and the charge on B remains unchanged

After switch S is opened, as the Capacitor A is connected across the battery, its potential difference is fixed at steady state (i.e., when capacitor is fully charged).

Capacitor B is isolated, so its charge gets fixed. But as we insert the dielectric, its capacitance changes, thus its potential difference also changes.

Option C is correct.