Consider two parallel-plate capacitors, C1 and C2, that are connected in paralle
ID: 1768607 • Letter: C
Question
Consider two parallel-plate capacitors, C1 and C2, that are connected in parallel. The capacitors are identical except that C2 has a dielectric inserted between its plates. A 210 V battery is connected across the combination until electrostatic equilibrium is established, and then the battery is disconnected.
(a) What is the charge on each capacitor? (Use the following as necessary: C1 and ?. Do not include the unit for the voltage in your answers.) Q1 =
Q2 =
(b) What is the total stored energy of the capacitors? (Use the following as necessary: C1 and ?. Do not include the unit for the voltage in your answers.)
U =
(c) The dielectric is removed from C2. What is the final stored energy of the capacitors? (Use the following as necessary: C1 and ?. Do not include the unit for the voltage in your answers.)
Uf =
(d) What is the final voltage across the two capacitors? (Use the following as necessary: C1 and ?. Do not include the unit for the voltage in your answers.)
Vf =
Explanation / Answer
a) Electric field strength is volts/m. Both capacitors have the same voltage but C2 has the smaller separation so it has the larger Electric field - its e) 2E
b) Its e) again! Twice the area doubles C and then half the distance doubles it again giving 4C as the capacitance of C2
c) a) this time. With 4 times the capacitance, C2 will hold 4 times the charge if its at the same voltage as C1 ( which it is.)