A dielectric slab is slowly inserted between the plates of a parallel plate capacitor while the capacitor is connected to a battery.As it is being inserted:
A)the capacitance, the potential difference between the plates, and the charge on the positive plate all increase
B)the capacitance, the potential difference between the plates, the charge on the positive plate all decrease
C)the potential difference between the plates increases, the charge on the positive plate decreases, and the capacitance remains the same
D)the capacitance and the charge on the positive plate decrease but the potential difference between the plates remains the same
E)the capacitance and the charge on the plate increase but the potential difference between the plates remains the same

Question

Quizplus · Accepted Answer

When a dielectric slab is inserted into a capacitor connected to a battery, the capacitance increases because the dielectric increases the capacitor's ability to store charge. The potential difference (voltage) across the plates remains constant because the battery maintains a constant voltage. As the capacitance increases, the charge on the plates also increases to maintain the constant voltage, according to the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.

A Dielectric Slab Is Slowly Inserted Between the Plates of a Parallel