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) To determine the potential difference between the end plates, we need to find the total capacitance of the system.

In a series combination of capacitors, the reciprocal of the total capacitance is equal to the sum of the reciprocals of the individual capacitances.

So, the total capacitance, C_total, can be calculated as:

1 / C_total = 1 / C1 + 1 / C2 + 1 / C3

Given that C1 = C2 = C3 = 120 pF, we can substitute the values into the equation:

1 / C_total = (1 / 120 pF) + (1 / 120 pF) + (1 / 120 pF)

1 / C_total = 3 / 120 pF

C_total = 120 pF / 3

C_total = 40 pF

Now, we can determine the potential difference, V, using the equation:

Q = C × V

where Q is the charge and C is the capacitance.

Q = C_total × V

Q = 40 pF × 500 V

Q = 20,000 pC

Therefore, the potential difference between the end plates is 500 V.

(b) To find the charge on each capacitor in a series combination, we need to divide the total charge, Q_total, among the capacitors.

Since the capacitors are in series, the charge remains the same for each capacitor. Thus, the charge on each capacitor is Q_total divided by the number of capacitors.

Q on each capacitor = Q_total / Number of capacitors

Q on each capacitor = 20,000 pC / 3

Q on each capacitor = 6,666.67 pC

Therefore, the charge on each capacitor is approximately 6,667 pC.

(c) The energy stored in a capacitor can be calculated using the equation:

U = 1/2 × C × V^2

where U is the energy, C is the capacitance, and V is the potential difference.

The energy stored in each capacitor is the same, but we need to find the total energy stored in the system. Since the capacitors are in series, the potential difference across each capacitor adds up to the total potential difference.

U_total = 1/2 × C_total × V^2

U_total = 1/2 × 40 pF × (500 V)^2

U_total = 1/2 × 40 pF × 250,000 V^2

U_total = 5000 pC × V^2

U_total = 5000 pC × (500 V)^2

U_total = 5000 pC × 250,000 V

U_total = 1.25 × 10^12 pCV

Therefore, the energy stored in the system is approximately 1.25 × 10^12 pCV.

To solve this problem, we can use the concept of capacitors in series. When capacitors are connected in series, the total capacitance (C_total) is given by the reciprocal of the sum of the reciprocals of individual capacitances (C_1, C_2, C_3):

1/C_total = 1/C_1 + 1/C_2 + 1/C_3

Let's calculate the total capacitance first:
1/C_total = 1/120 pF + 1/120 pF + 1/120 pF
1/C_total = 3/120 pF
C_total = 120 pF/3
C_total = 40 pF

Now, let's go through the questions one by one:

(a) The potential difference between the end plates of the capacitors connected in series is the sum of the potential differences across each individual capacitor. Since each capacitor has the same charge (Q) but different capacitance (C), the potential difference across each capacitor can be calculated using the formula:

V = Q/C

For the given capacitors, since they have the same charge and the same capacitance, the potential difference across each capacitor is:

V = Q/120 pF

Now, substitute the known values to calculate the potential difference:

V = 500 V / 120 pF
V = 4.16 V (rounded to two decimal places)

Therefore, the potential difference between the end plates is 4.16 V.

(b) Since the capacitors are connected in series, the charge on each capacitor is the same. Let's denote the charge on each capacitor as Q_total.

The total charge (Q_total) is given by the formula:

Q_total = C_total * V_total

Substituting the known values:

Q_total = 40 pF * 4.16 V
Q_total = 166.4 pC (rounded to one decimal place)

Therefore, the charge on each capacitor is 166.4 picocoulombs.

(c) The energy stored in the system can be calculated using the formula:

E = (1/2) * C_total * V_total^2

Substituting the known values:

E = (1/2) * 40 pF * (4.16 V)^2
E = 346.75 pJ (rounded to two decimal places)

Therefore, the energy stored in the system is 346.75 picojoules.