Final answer:
Upon inserting a ceramic dielectric with a constant of k = 5.8 into a 2800-pF air-filled capacitor connected to a 16-V battery, a charge of 215.04 μC will flow from the battery.
Step-by-step explanation:
The question is about calculating the charge that will flow from a battery when a dielectric material is inserted into an air-filled capacitor connected to it.
To find out how much charge will flow, one must consider the initial and final capacitance of the capacitor and use the relationship Q = C × V, where Q is the charge, C is the capacitance, and V is the potential difference (voltage).
Initially, the capacitor has a capacitance of 2800 pF and is connected to a 16-V battery, so the initial charge (Qinitial) is given by:
Qinitial = Cinitial × V
Qinitial = 2800 pF × 16 V
Qinitial = 44800 pF×V
When the dielectric is inserted, the capacitance increases by a factor of the dielectric constant (k = 5.8). Therefore, the new capacitance (Cfinal) is:
Cfinal = Cinitial × k
Cfinal = 2800 pF × 5.8
Cfinal = 16240 pF
Since the capacitor is still connected to the 16-V battery, the final charge (Qfinal) will be:
Qfinal = Cfinal × V
Qfinal = 16240 pF × 16 V
Qfinal = 259840 pF×V
To find the amount of charge that will flow from the battery, we subtract the initial charge from the final charge:
ΔQ = Qfinal - Qinitial
ΔQ = 259840 pF×V - 44800 pF×V
ΔQ = 215040 pF×V
Converting to coulombs (since 1 pF = 10-12 F):
ΔQ = 215040 × 10-12 C/V × V
ΔQ = 215.04 × 10-6 C
ΔQ = 215.04 μC
Therefore, a charge of 215.04 μC will flow from the battery when the ceramic dielectric material is inserted into the air-filled capacitor.