Answer:
A. The new capacitance is half in term of C
B. The charge remains the same in terms of [tex]Q_{0}[/tex]
C. The potential difference is double in terms of [tex]V_{0}[/tex]
Explanation:
The battery with a voltage of [tex]V_{0}[/tex] is used to charge the plates, giving it a capacitance of C.
The charging process leaves a charge of magnitude [tex]Q_{0}[/tex] on the plate
The battery is disconnected (this will leave it with a constant charge [tex]Q_{0}[/tex])
the relationship between the charge, voltage and capacitance of the plate is
[tex]Q_{0}[/tex] = C[tex]V_{0}[/tex] ......... equ 1
A. The relationship between capacitance and the distance of the plate is given as
C = Aε/d ......... equ 2
where A is the area of the plate,
ε is the permeability of free space,
d is the distance between the plates
The area of the plate does not change, and permeability of free space is a constant. The combination of all these means that if the distance is doubled, then the capacitance will be halved. This is from equ 2 when the distance becomes 2d, then we have
C' = Aε/2d
==> C' = C/2
B. Since the battery is disconnected, and the capacitor is not discharged, the charge on the plate will remains the same as [tex]Q_{0}[/tex]. This is due to the conservation of charges.
C. Since the charge remains constant, and the capacitance is halved, then from equ 1, the new potential difference V will become double of the initial potential difference [tex]V_{0}[/tex]
==> V = 2[tex]V_{0}[/tex]
