A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn

(a) 12.0 V

The capacitor remains connected to the 12.0 V source until fully charged. Taking into account the capacitor's capacity, , the total charge held at the conclusion of this process is

Upon disconnecting the battery, the charge on the capacitor stays the same. The capacitance, C, does not change either, as it solely relies on the capacitor's characteristics (area and distance between plates), which stay constant. Hence, according to the relationship

and since neither Q nor C vary, the voltage V also remains at 12.0 V.

(b) (i) 24.0 V

In this scenario, the distance between the plates is doubled. Let's recall the formula for the capacitance of a parallel-plate capacitor:

where:

is the permittivity of free space

is the relative permittivity of the capacitor's material

A is the area of the plates

d is the distance separating the plates

As noted, the plate distance is now doubled: d'=2d. This indicates that the capacitance is halved: . The updated voltage across the plates is given by

Since Q (the charge) remains constant (the capacitor is isolated, preventing charge flow), the new voltage turns out to be

This leads us to a new voltage of

(c) (ii) 3.0 V

The area for each capacitor plate is determined as:

where r indicates the radius of the plate. In this situation, the radius is doubled: r'=2r. Therefore, the new area becomes

While the distance between the plates remained unchanged (d); consequently, the new capacitance can be expressed as

Thus, the capacitance has quadrupled, leading to a new voltage of

which equates to