A 62 V battery charges a 3875 pF air gap capacitor. After disconnecting, paraffin is placed between the plates. What is the resulting charge of the plates?
1.23 x 10-6 C
1.73 x 10-6 C
1.54 x 10-6 C
1.44 x 10-6 C
I swear I was doing it right but I keep getting the wrong answer, can someone help me solve this?
derfram
Favorite Answer
Q=CV
Q = 3875 pF * 62 V = 2.4025 x 10^-7 C
Adding the paraffin will not change the amount of charge. It will change the value of capacitance, and the voltage, but the charge will remain the same.
2.4025 x 10^-7 C. None of the suggested answers are correct.
oldschool
The charge can't increase or decrease since you have disconnected the battery. The capacitance changes and V = Q/C