2 parallel plates with area .1m^2 at distance of .071 m. Capacitor is connected
ID: 1327476 • Letter: 2
Question
2 parallel plates with area .1m^2 at distance of .071 m. Capacitor is connected to 3.7 V battery then disconnected. Distance between plates changes to .063 m and then connected to 5.7 V battery. What amount of charge flows from capacitor to the new battery? 2 parallel plates with area .1m^2 at distance of .071 m. Capacitor is connected to 3.7 V battery then disconnected. Distance between plates changes to .063 m and then connected to 5.7 V battery. What amount of charge flows from capacitor to the new battery? 2 parallel plates with area .1m^2 at distance of .071 m. Capacitor is connected to 3.7 V battery then disconnected. Distance between plates changes to .063 m and then connected to 5.7 V battery. What amount of charge flows from capacitor to the new battery?Explanation / Answer
With 1st battery:
C= ebsoleneo*A/d
=(8.854*10^-12* 0.1) / (0.071)
=1.25*10^-11 F
Q= C*V
=(1.25*10^-11) * 3.7
= 4.6*10^-11 C
With 2nd battery:
C= ebsoleneo*A/d
=(8.854*10^-12* 0.1) / (0.063)
=1.41*10^-11 F
Q= C*V
=(1.41*10^-11) * 5.7
= 8.01*10^-11 C
Charge will flow fromnew battery to capacitor.
This is:
8.01*10^-11 - 4.6*10^-11 =3.4 *10^-11 C
Answer: 3.4 *10^-11 C
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