Four capacitors are connected in series with a battery, as in the figure below, where C1 = 3.24 µF, C2 = 6.11 µF, C3 = 12.1 µF, C4 = 25.0 µF, V = 18.4 V.

(a) Calculate the capacitance of the equivalent capacitor.
µF

(b) Compute the charge on C3.
µC

(c) Find the voltage drop across C3.
V

C4 is removed from the circuit, leaving only three capacitors in series.
(a) Find the equivalent capacitance.
Ceq = µF

(b) Find the charge on C2.
Q = µC

(c) Find the voltage drop across C2.
ΔV = V

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An electron traveling horizontally from North to South enters a region where a uniform electric field is directed downward. What is the direction of the electric force exerted on the electron once it has entered the field?

(a) To find the capacitance of the equivalent capacitor, Ceq, when capacitors are connected in series, we use the formula:

1/Ceq = 1/C1 + 1/C2 + 1/C3 + 1/C4

Substituting the given values, we get:

1/Ceq = 1/3.24µF + 1/6.11µF + 1/12.1µF + 1/25.0µF

Calculating this expression will give us the reciprocal of the equivalent capacitance, so we take the reciprocal of the final result to find the value of Ceq.

1/Ceq = 0.3086 + 0.1638 + 0.0826 + 0.04

1/Ceq = 0.595

Ceq = 1/0.595

Ceq ≈ 1.680 µF

Therefore, the capacitance of the equivalent capacitor is 1.680 µF.

(b) To find the charge on C3, we can use the formula:

Q = C3 * V

Q = 12.1µF * 18.4V

Calculating this expression:

Q = 222.04 µC

Therefore, the charge on C3 is approximately 222.04 µC.

(c) The voltage drop across C3 can be found using the formula:

ΔV = Q / C3

ΔV = 222.04 µC / 12.1µF

Calculating this expression:

ΔV ≈ 18.28V

Therefore, the voltage drop across C3 is approximately 18.28V.

When C4 is removed from the circuit:

(a) The equivalent capacitance, Ceq, is given by the formula:

1/Ceq = 1/C1 + 1/C2 + 1/C3

Substituting the given values:

1/Ceq = 1/3.24µF + 1/6.11µF + 1/12.1µF

Calculating this expression:

1/Ceq = 0.3086 + 0.1638 + 0.0826

1/Ceq = 0.555

Ceq = 1/0.555

Ceq ≈ 1.800µF

Therefore, the equivalent capacitance is approximately 1.800 µF.

(b) The charge on C2 can be found using the equation:

Q = Ceq * V

Q = 1.800µF * 18.4V

Calculating this expression:

Q = 33.12 µC

Therefore, the charge on C2 is approximately 33.12 µC.

(c) The voltage drop across C2 can be found using the formula:

ΔV = Q / C2

ΔV = 33.12µC / 6.11µF

Calculating this expression:

ΔV ≈ 5.41V

Therefore, the voltage drop across C2 is approximately 5.41V.

To answer these questions, we need to apply the rules for capacitors connected in series.

The total capacitance (Ceq) for capacitors connected in series is given by the reciprocal of the sum of the reciprocals of individual capacitances.

(a) To calculate the capacitance of the equivalent capacitor, we substitute the given values into the formula:

1/Ceq = 1/C1 + 1/C2 + 1/C3 + 1/C4

Substituting the values:
1/Ceq = 1/3.24µF + 1/6.11µF + 1/12.1µF + 1/25.0µF

Then, we can calculate Ceq by taking the reciprocal of both sides of the equation:

Ceq = 1 / (1/3.24µF + 1/6.11µF + 1/12.1µF + 1/25.0µF)

(b) To compute the charge on C3, we can use the equation:

Q = C * V

Substituting the given values:
Q = 12.1µF * 18.4V

(c) To find the voltage drop across C3, we can use the formula:

V = Q / C

Substituting the known values:
V = Q / 12.1µF

Now, let's consider the new situation where C4 is removed from the circuit, and only three capacitors remain in series.

(a) To find the equivalent capacitance (Ceq) in this case, we use the same formula:

1/Ceq = 1/C1 + 1/C2 + 1/C3

Substituting the given values:
1/Ceq = 1/3.24µF + 1/6.11µF + 1/12.1µF

Taking the reciprocal to solve for Ceq:

Ceq = 1 / (1/3.24µF + 1/6.11µF + 1/12.1µF)

(b) To determine the charge on C2, we can again use the formula:

Q = C * V

Substituting the relevant values:
Q = 6.11µF * 18.4V

(c) Finally, to find the voltage drop across C2, we can use the equation:

V = Q / C

Substituting the known values:
V = Q / 6.11µF