Three capacitors, each of capacitance 120 pF, are each charged to 500 V and then connected in series. Determine (a) the potential difference between the end plates, (b) the charge on each capacitor, and (c) the energy stored in the system
a. E = 3 * 500 = 1500 Volts.
b. Ceq = 120/3 = 40 pF.
Q = Ceq*E = 40 * 1500 = 60,000 Pico Coulombs.
c. Energy = 0.5Ceq*E^2 = 0.5*40 * 1500^2 =
To find the potential difference between the end plates of the series capacitors, you need to use the formula for adding capacitors in series:
1/Ceq = 1/C1 + 1/C2 + 1/C3 + ...
In this case, since all three capacitors have the same capacitance of 120 pF, the equation becomes:
1/Ceq = 1/120 pF + 1/120 pF + 1/120 pF
Simplifying this equation yields:
1/Ceq = 3/120 pF = 1/40 pF
Now, take the reciprocal of both sides of the equation to find the equivalent capacitance (Ceq):
Ceq = 40 pF
The potential difference across capacitors in series remains the same. Therefore, the potential difference between the end plates of the series capacitors is 500 V.
To determine the charge on each capacitor, you can use the formula:
Q = C * V
Since all three capacitors have the same capacitance (C) and potential difference (V), the charge on each capacitor is:
Q1 = Q2 = Q3 = (120 pF) * (500 V) = 60,000 pC
To find the energy stored in the system, you can use the formula:
E = (1/2) * C * V^2
Since all three capacitors have the same capacitance (C) and potential difference (V), the energy stored in the system is:
E = (1/2) * (40 pF) * (500 V)^2 = 5,000,000 pJ
(a) The potential difference between the end plates is 500 V.
(b) The charge on each capacitor is 60,000 pC.
(c) The energy stored in the system is 5,000,000 pJ.