Capacitors in Parallel problems and solutions

 Problem #1

How many 1.00 μF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the capacitors?

Answer;
Known:
Capacitances N capasitor is C =  1.00 μF, with N is the number of capacitors
potential across the capacitors is V = 110 V
charge, q  = 1.00 C
Hence the number of parallel arranged capacitors is given by

NC = q/V
N = q/VC = 1.00 C/(1.00 x 10^─6 F x 110 V) = 9091 capacitors

Problem #2
Each of the uncharged capacitors in Fig. 01 has a capacitance of 25.0 μF. A potential difference of V = 4200 V is established when the switch is closed. How many coulombs of charge then pass through meter A?

Fig.01

Answer;
Capacitors connected in parallel can be replaced with an equivalent capacitor that has the same total charge q and the same potential difference V as the actual capacitors.

The total capasitance C_P of the parallel caacitors of  C given by

C_P = 3C

C_P = 3 x 25.0 μF = 75 μF

Then

q_A = C_PV = (75μF)(4200 V) = 315 mC

Problem #3
Figure 02 shows a variable “air gap” capacitor for manual tuning. Alternate plates are connected together; one group of plates is fixed in position, and the other group is capable of rotation. Consider a capacitor of n = 8 plates of alternating polarity, each plate having area A = 1.25 cm² and separated from adjacent plates by distance d = 3.40 mm. What is the maximum capacitance of the device?

Fig.02

Answer:
Known:
number of identical capacitors, n = 8
each plate having area A = 1.25 cm² = 1.25 x 10^-4 m²
distance d = 3.40 mm = 3.40 x 10^-3 m

For maximum capacitance the two groups of plates must face each other with maximum area. In this case the whole capacitor consists of (n – 1) identical single capacitors connected in parallel. Each capacitor has surface area A and plate separation d so its capacitance is given by

C₀ = e₀A/d.

Thus, the total capacitance of the combination is

C = (n – 1)C₀ = (n – 1)(e₀A/d)

   = (8 – 1)(8.85 x 10^-12 C²/N.m²)(1.25 x 10^-4 m²)/(3.40 x 10^-3 m)

C = 2.72 x 10^-12 F = 2.71 μF

Problem #4
In Fig. 03, the capacitances are C₁ = 1.0 μF and C₂ = 3.0 μF, and both capacitors are charged to a potential difference of V = 100 V but with opposite polarity as shown. Switches S₁ and S₂ are now closed. (a) What is now the potential difference between points a and b? What now is the charge on capacitor (b) 1 and (c) 2?

Fig.03

Answer:
The charge on each cap. before switches are closed is;

Q₁ = C₁V₀ = 100 μF x 100 V = 10⁴ μC
Q₂ = C₂V₀ = 300 μF x 100 V =  3 x 10⁴ μC

When the switches are closed the charge redistributes into q₁ and q₂ but the total charge is less because of the initial reverse polarity. The total is;

q₁ + q₂ = Q₂ – Q₁

C₁V + C₂V = C₂V₀ - C₁V₀

V = (C₂ – C₁)V₀/(C₂ + C₁)

   = (3.0 μF – 1 0 μF)100 V/(3.0 μF + 1.0 μF)

V = 50 Volt

This is the final voltage across either capacitor (they are in parallel) which is also the voltage V_ab, so

V_ab = 50 Volt

 (b) the charge on capacitor 1 is

Q₁ = C₁V₀ = 100 μF x 50 V = 5 x 10³ μC = 5 mC

 (b) the charge on capacitor 2 is

Q₁ = C₁V₀ = 300 μF x 50 V = 15 x 10³ μC = 15 mC

Problem #5
In Fig. 04, V = 10 V, C₁ = 10 μF, and C₂ = C₃ = 20 μF. Switch S is first thrown to the left side until capacitor 1 reaches equilibrium. Then the switch is thrown to the right. When equilibrium is again reached, how much charge is on capacitor 1?

Answer:
Switch to left

Initial charge on C₁

Q₁ = C₁V₀ =  10 μF x 10 V = 100 μF

capacitor C₂ and C₃ are in parallel

C₂₃ = C₂ + C₃ = 40 μF

Switch to right

Charge is conserved:

Q = Q₁ + Q₂₃

C₁V₀ = C₁V₁ + C₂₃V₂₃ with V₁ = V₂₃ = V, then

C₁V₀ = C₁V + C₂₃V

V = C₁V₀/( C₁ + C₂₃) = (10 μF)(10 V)/(10 μF + 40 μF) = 2.0 V = V₁ = V₂₃
So that

Q₁ = C₁V = 10 μF x 2.0 V = 20 μC and

Q₂₃ = C₂₃V = 40 μF x 2.0 V = 80 μC

Problem #6
In Fig. 05, two parallel-plate capacitors (with air between the plates) are connected to a battery. Capacitor 1 has a plate area of 1.5 cm² and an electric field (between its plates) of magnitude 2000 V/m. Capacitor 2 has a plate area of 0.70 cm² and an electric field of magnitude 1500 V/m. What is the total charge on the two capacitors?

Answer:
Known:
Capacitor 1 has a plate area, A₁ = 1.5 cm² = 1.5 x 10^─4 m²
electric field in capacitor 1, E₁ = 2000 V/m
Capacitor 2 has a plate area, A₂ = 0.7 cm² = 7 x 10^─5 m²
electric field in capacitor 1, E₂ = 1500 V/m

the capacitance of each capacitor is

C₁ = ϵ₀A₁/d₁

C₂ = ϵ₀A₂/d₂

The relationship between the electric field and electric potential is given by
E = V/d, then V = Ed

the electric charge on each capacitor is

Q₁ = C₁V =  (ϵ₀A₁/d₁)E₁d₁ = ϵ₀A₁E₁ and

Q₂ = C₂V =  (ϵ₀A₂/d₂)E₂d₂ = ϵ₀A₂E₂

then, the total charge on the capacitor is

Q = Q₁ + Q₂ = ϵ₀(A₁E₁ + A₂E₂)

Q = 8.85 x 10^-12 C²/N.m²[(1.5 x 10^─4 m² x 2000 V/m + (0.7 x 10^─4 m² x 1500 V/m)

Q = 3.58 pC    

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