PHYS 1220 Capacitors Solutions

Capacitor Solutions

In the figure below, C₁ = C₅ = 3.00 μF and C₂ = C₃ = C₄ = 2.00 μF . What is the equivalent capacitance of the circuit? A potential of 600 V is applied across points A and B. What is the charge on each capacitor? What is the voltage across each capacitor? What is the energy stored in each capacitor? Put your results in a table.

We reduce a capacitor circuit to its equivalent by reducing portions of the circuit one at a time. Note, capacitors C₃ and C₄ are in series and therefore 1/C₃₄ = ½ + ½ = 1, or C₃₄ = 1 μF. A potential of 24 V is applied across points A and B. What is the charge on each capacitor? What is the voltage across each capacitor? What is the energy stored in each capacitor? Put your results in a table

Capacitors C₂ and C₃₄ are in parallel and therefore C₂₃₄ = 2 + 1 = 3 μF.

Final C₁ and C₂₃₄ and C₅ are is series and therefore 1/C₁₂₃₄₅ = 1/3 + 1/3 +1/3 = 1, or C₁₂₃₄₅ = 1 μF. The equivalent capacitance is 1 μF.

Once we are down to the equivalent capacitor we can use the relationships that Q = CV and U = ½CV and the following facts

*Series*	*Parallel*
1/C_s = 1/C₁ + 1/C₂	C_p = C₁ + C₂
Same charge on C_s, C₁, and C₂	Same Voltage
Different voltage	Different charge on C_p, C₁, and C₂

to find the Q, V, and U for each capacitor as we work back up to the original circuit.

(a) Since V = 600 Volts and C₁₂₃₄₅ = 1 μF, the charge on the equivalent capacitor is Q₁₂₃₄₅ = 600 μC. The energy is U₁₂₃₄₅ = ½QV = 0.180 J.

(b) The capacitor C₁₂₃₄₅ is actually C₁, C₂₃₄, and C₅ in series, so each has the same charge of 600 μC. The voltage drop over each is V₁ = V₂₃₄ = V₅ = Q/C = 600 μC / 3 μF = 200 V. The energy for each using U = ½QV yields U₁ = U₂₃₄ = U₅ = 0.060 J.

(c) The capacitor C₂₃₄ is actually C₂ and C₃₄ in parallel, so there is 200 volts over each. The charge on C₂ is Q₂ = C₂V = 400 μC. The charge on C₃₄ is Q₃₄ = C₃₄V = 200 μC. The energy for each using U = ½QV yields U₂ = 0.040 J and U₃₄ = 0.020 J.

A potential of 24 V is applied across points A and B. What is the charge on each capacitor? What is the voltage across each capacitor? What is the energy stored in each capacitor? Put your results in a table

(d) The capacitor C₃₄ is actually C₃ and C₄ in series, so each has the same charge of 200 μC. The voltage drop over each is V₂ = V₄ = Q/C = 200 μF / 2 μF = 100 V. The energy for each using U = ½QV yields U₂ = U₄ = 0.010 J.

Capacitor	Capacitance	Voltage	Charge	Energy
C₁	3.00 μF	200 V	600 μC	0.060 J
C₂	2.00 μF	200 V	400 μC	0.040 J
C₃	2.00 μF	100 V	200 μC	0.010 J
C₄	2.00 μF	100 V	200 μC	0.010 J
C₅	3.00 μF	200 V	600 μC	0.060 J

What is the equivalent capacitance of the circuit shown below? A potential of 24 V is applied across points A and B. What is the charge on each capacitor? What is the voltage across each capacitor? What is the energy stored in each capacitor? Put your results in a table.

The 1 μF and 3 μF capacitors are in parallel and are equivalent to a single 4 μF capacitor. The 6 μF and 2 μF capacitors are also in parallel and are equivalent to a single 8 μF capacitor.

The two 4 μF capacitors are in series, therefore 1/C_p = ¼ + ¼ = ½, or C_p = 2 μF. The two 8 F capacitors are in series, therefore 1/C_p = 1/8 + 1/8 = ¼, or C_p = 4 μF.

The 2 μF and 4 μF capacitors are in parallel and are equivalent to a single 6 μF capacitor. The equivalent capacitance is 6 μF.

Once we are down to the equivalent capacitor we can use the relationships that Q = CV and U = ½CV and the following facts

*eries*	*Parallel*
1/C_s = 1/C₁ + 1/C₂	C_p = C₁ + C₂
Same charge on C_s, C₁, and C₂	Same Voltage
Different voltage	Different charge on C_p, C₁, and C₂

to find the Q, V, and U for each capacitor as we work back up to the original circuit.

(a) Then we work backwards from the equivalent capacitor. Since V = 24 Volts and C_eq = 6 μF, the charge on the equivalent capacitor is Q_eq = 24 V × 6 μF = 144 μC. The energy is U_eq = ½QV = 1.728 mJ.

(b) The equivalent capacitor is actually a 2 μF and a 4 μF capacitor in parallel, so each has the same voltage of 24 V. The charge on the 2 μF capacitor is Q = 2μF × 24 V = 48 μC. The charge on the 4 μF capacitor is Q = 4μF × 24 V = 96 μC. The energy for each using U = ½QV yields U₂ = 0.576 mJ and U₄ = 1.152 mJ.

(c) The 2 μF capacitor is actually two 4 μF capacitors in series, so each has the same charge of 48 μC. The voltage drop over each is V = Q/C = 48 μC / 4 μF = 12 V. The energy for each using U = ½QV yields U_4R = U_4L = 0.288 mJ.

The 4 μF capacitor is actually two 8 μF capacitors in series, so each has the same charge of 96 μC. The voltage drop over each is V = Q/C = 96 μC / 8 μF = 12 V. The energy for each using U = ½QV yields U_8R = U_8L = 0.576 mJ.

(d) The top right 4F capacitor is actually a 1 μF and a 3 μF capacitor in parallel, so each has the same voltage of 12 V. The charge on the 1 μF capacitor is Q = 1μF × 12 V = 12 μC. The charge on the 3 μF capacitor is Q = 3μF × 12 V = 36 μC. The energy for each using U = ½QV yields U₁ = 0.072 mJ and U₃ = 0.216 mJ.

The bottom left 8F capacitor is actually a 6 μF and a 2 μF capacitor in parallel, so each has the same voltage of 12 V. The charge on the 6 μF capacitor is Q = 6μF × 12 V = 72 μC. The charge on the 2 μF capacitor is Q = 2μF × 12 V = 24 μC. The energy for each using U = ½QV yields U₆ = 0.432 mJ and U₂ = 0.144 mJ.

Capacitor	Voltage	Charge	Energy
4 μF	12 V	48 μC	0.288 mJ
1 μF	12 V	12 μC	0.072 mJ
3 μF	12 V	36 μC	0.216 mJ
8 μF	12 V	96 μC	0.576 mJ
6 μF	12 V	72 μC	0.432 mJ
2 μF	12 V	24 μC	0.144 mJ

A 4.00 μF capacitor and an 8.00 μF capacitor are separately charged by a 20.0 Volt power supply. The capacitors are then placed in the circuit shown below.

(a) What is the charge on each capacitor when switch S is open?
(b) What will be the charge and polarity of each capacitor when switch S is closed?

(a) The charge on the 4 μF capacitor is

Q₄ = CV = (4 μF)(20 V) = 80 μC.

The charge on the 8 μF capacitor is

Q₈ = CV = (8 μF)(20 V) = 160 μC.

(b) Capacitors in parallel must have the same voltage drop (size and direction). Here the voltage drops are opposite, so charge must flow until the voltage drop is the same.

Assume positive charge q moves to the 4 μF capacitor from the 8 μF capacitor but that the polarity is unchanged as we move around the circuit. This means the new charge is 80 μC + q on the 4 μF capacitor and 160 μC + q on the 8 μF capacitor where we have conserved charge and been mindful of the polarity. Kirchhoff’s voltage rule yields

(80 μC + q) / 4 μF + (160 μC + q) / 8 μF = 0.

Rearranging to get q by itself

(80 μC / 4 μF + 160 μC / 8 μF) + q (1 / 4 μF + 1 / 8 μF) = 0.

Which is the same as

(20 V + 20 V) + q (3 / 8 μF) = 0.

Solving for q yields q = –320/3 μC. So there is now 80 μC + q = –80/3 μC on the 4 μF capacitor and V₄ = Q/C = –6²/₃ V. Then minus sign indicates that the polarity of the 4 μF capacitor has changed. On the 8 μF capacitor, the new charge is 160 μC + q = 160/3 μC and V₈ = Q/C = 6²/₃ V.

Note that at the start of the problem the total energy was U₁ + U₂ = ½(4 μF)(20 V) + ½(8 μF)(20 V) = 40 μJ + 80 μJ = 120 μJ. At the end of the problem, the total energy s U_1f + U_2f = ½(4 μF)(6²/₃ V) + ½(8 μF)(6²/₃ V) = 13¹/₃ μJ + 26²/₃ μJ = 40 μJ. Energy is not conserved because we are treating these capacitors as ideal. Real capacitors have resistance and there is Joule Heating by the current through this resistance.

A 4.00 μF capacitor and an 8.00 μF capacitor are separately charged by a 20.0 Volt power supply. The capacitors are then placed in the circuit shown below. What will be the charge and polarity of each capacitor when both switches are closed? (Note that the polarities are switched from the previous question.)

When we also close both switches, the two capacitors are connected in parallel. Capacitors in parallel must have the same voltage drop (sizeA capacitor consists o and direction). Here the voltage drops are the same, unlike the previous question, so no charge will flow.

If you like you can assume that positive charge q moves to the 4 μF capacitor from the 8 μF capacitor but that the polarity is unchanged as we move around the circuit. This means the new charge is 80 μC + q on the 4 μF capacitor and 160 μC – q on the 8 μF capacitor where we have conserved charge and been mindful of the polarity. Kirchhoff’s voltage rule yields

(80 μC + q) / 4 μF – (160 μC – q) / 8 μF = 0.

Rearranging to get q by itself

(80 μC / 4 μF – 160 μC / 8 μF) + q (1 / 4 μF + 1 / 8 μF) = 0.

Which is the same as

(20 V – 20 V) + q (3 / 8 μF) = 0.

Solving for q yields q = 0. Charge does not move.

A 10 μF capacitor is charged to 10 V while a 20 μF capacitor is charged to 5 V. They are then placed in the circuit below with given orientation. What will be the charge and voltage drop across each capacitor when the switch is closed?

Since the sum of the voltages is not zero around the circuit, charge will flow until Kirchhoff's voltage rule is obeyed. The initial charge on the 10 μF capacitor is Q₁₀ = CV = 10 μF × 10 V = 100 μC while the charge on the 20 μF capacitor is Q₂₀ = CV = 20 μF × 5 V = 100 μC. Next assume positive charge q moves to the 10 μF capacitor from the 20 μF capacitor but that the polarity is unchanged as we move around the circuit. This means the new charge is 100 μC + q on the 10 μF capacitor and 100 μC – q on the 20 μF capacitor where we have conserved charge and been mindful of the polarity. Kirchhoff’s voltage rule yields

(100 μC + q) / 10 μF – (100 μC – q) / 20 μF = 0.

Rearranging to get q by itself

(100 μC / 10 μF – 100 μC / 20 μF) + q (1 / 10 μF + 1 / 20 μF) = 0.

Which is the same as

(10 V – 5 V) + q (3 / 20 μF) = 0.

Solving for q yields q = –100/3 μC. So there is only 100 μC + q = 200/3 μC on the 10 μF capacitor and V₁₀ = Q/C = 6²/₃ V. On the 20 μF capacitor, the new charge is 100 μC – q = 400/3 μC and V₂₀ = Q/C = 6²/₃ V.

Note that at the start of the problem the total energy was U₁ + U₂ = ½(10 μF)(10 V) + ½(20 μF)(5 V) = 50 μJ + 50 μJ = 100 μJ. At the end of the problem, the total energy is U_1f + U_2f = ½(10 μF)( 6²/₃ V) + ½(20 μF)( 6²/₃ V) = 33¹/₃ μJ + 66²/₃ μJ = 100 μJ.

Since the sum of the voltages is not zero around the circuit, charge will flow until Kirchhoff’s voltage rule is obeyed. The initial charge on the 10 μF capacitor is Q₁₀ = CV = 10 μF × 10 V = 100 μC while the charge on the 20 μF capacitor is Q₂₀ = CV = 20 μF × 5 V = 100 μC as shown in the previous question. Next assume positive charge q moves to the 10 μF capacitor from the 20 μF capacitor and that the polarity is unchanged as we move around the circuit. This means the new charge is 100 μC + q on the 10 μF capacitor and 100 μC + q on the 20 μF capacitor where we have conserved charge and been mindful of the polarity. Applying Kirchhoff’s voltage rule yields

(100 μC + q) / 10 μF + (100 μC + q) / 20 μF = 0.

Rearranging to get q by itself

(100 μC / 10 μF + 100 μC / 20 μF) + q (1 / 10 μF + 1 / 20 μF) = 0.

Which is the same as

(10 V + 5 V) + q (3 / 20 μF) = 0.

Solving for q yields q = –100 μC. So there is 100 μC + q = 0 μC finally on the 10 μF capacitor and V₁₀ = Q/C = 0 V. On the 20 μF capacitor, the new charge is 100 μC + q = 0 μC and V₂₀ = Q/C = 0 V. The capacitors have discharged themselves

Note that at the start of the problem the total energy was U₁ + U₂ = ½(10 μF)(10 V) + ½(20 μF)(5 V) = 50 μJ + 50 μJ = 100 μJ. At the end of the problem, the total energy is U_1f + U_2f = 0. Energy is not conserved because we are treating these capacitors as ideal. Real capacitors have resistance and there is Joule Heating by the current through this resistance.

Two oppositely charged conducting plates, with equal magnitude of charge per unit area, are separated by a dielectric 3.00 mm thick, with a dielectric constant of 4.50. The resultant electric field in the dielectric is 1.60 × 10⁶ V/m. Compute (a) the charge per unit area on the conducting plates, and (b) the charge per unit area on the surfaces of the dielectric.

(a) The relationship between the charge density on the capacitor plates and the field between them is E = E₀/κ = σ_free/κε₀ , where E is the electric field with the dielectric in place and E₀ is the field when there is no dielectric. Therefore the charge density is

σ_free = Eκε₀ = (1.60 × 10⁶ V/m)(4.50)(8.85 × 10^-12μF/m) = 63.72 μC/m² .

(b) The bound charge density is related to the free charge density by

σ_bound = - [(κ-1)/κ] σ_free = -49.56 μC/m² .

The minus sign indicates that the dielectric charge is opposite to the plate charge.

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is 3.60 × 10⁵ V/m. When the space is filled with a dielectric, the electric field is 1.20 × 10⁵ V/m.

(a) What is the charge density on the surface of the dielectric?
(b) What is the dielectric constant?

(a) With no dielectric, the relationship between the charge density on the capacitor plates and the field between them is E₀ = σ_free/ε₀ , so the charge density is

σ_free = E₀ε₀ = (3.60 × 10⁵ V/m)(8.85 × 10^-12 μF/m) = 3.186 μC/m² .

With the dielectric, the relationship between the net charge density on the capacitor plates and on the dielectric and the field between them is E = σ_net/ε₀ , so the charge density is

σ_net = E₀ = (1.20 × 10⁵ V/m)(8.85 × 10^-12 μF/m) = 1.062 μC/m² .

Since the net charge density is the sum of the free and bound charge densities, we find

σ_bound = σ_net - σ_free = 1.062 μC/m² - 3.186 μC/m² = -2.124 μC/m² .

The minus sign indicates that the dielectric charge is opposite to the plate charge.

(b) The dielectric constant is given by

κ = E₀/E = (3.60 × 10⁵ V/m)/(1.20 × 10⁵ V/m) = 3.00 .

A capacitor that has air between its plates is connected across a potential difference of 12 V and stores 48 μC of charge. It is then disconnected from the source while still charged. (a) find the capacitance of the capacitor. (b) A piece of Teflon (κ = 2.1) is inserted between the plates. find the capacitance of, the voltage across, and charge on the capacitor.

(a) The capacitance is given by

C₀ = Q/V = 48 μC / 12 V = 4 μF .

(b) Since the charge on the plates has no possible path from the plate, it is unchanged. The capacitance becomes C = κC₀ = 8.4 μF. The voltage across the capacitor will be

V = Q/C = 48 μC / 8.4 μF = 5.71 Volts.

Determine (a) the capacitance and (b) the maximum voltage that can be applied to a Teflon-filled (κ = 2.1) parallel-plate capacitor having a plate area of 1.75 cm² and dielectric thickness of 0.04 mm. Breakdown occurs when the electric field exceeds 60 × 10⁶ V/m.

(a) The capacitance of a parallel plate capacitor is given by

C = κε₀A/d = (2.1)(8.85 × 10^-12 F/m)( 1.75 × 10^-4 m²)/(0.04 × 10³ m) = 81.31 pF .

(b) From the electric field at breakdown we can find the charge density on the plates,

E = σ_free/κε₀,

σ_free = κε₀E .

The voltage across the capacitor is related to the charge by

V = Qσ_free/C = σ_free/(C/A) = dε_free/κε₀ .

Combining our results yields

V = Ed = (0.04 × 10^-3 m)(60 × 10⁶ V/m) = 2400 Volts .

Three 10-μF capacitors are connected in parallel. A dielectric κ = 2.0 is inserted into one of the capacitors. The capacitors are then connected to a 4.0 V battery. (a) What is the charge on each capacitor and what is the energy stored by each capacitor? (b) The battery is disconnected. The dielectric is then removed from the capacitor. What is the new charge on each capacitor and what now is the energy stored by each capacitor? (c) Compare the energy of the system before and after the dielectric is removed. Where did this energy come from and why?

(a) The presence of the dielectric gives the third capacitor a capacitance of 20 μF. Since the capacitors are in parallel, the voltage drop is 4 volts over each. The charge on each is given by Q=CV. So the charge on the top and middle capacitor is Q_top = Q_middle = 10 μF × 4 V = 40 μC. The charge on the bottom capacitor is Q_bottom = 20 μF × 4 V = 80 μF. The energy is given by U = ½CV². Hence the energy stored in the top and middle capacitor is U_top = U_middle = ½(10 μF)(4 V)² = 80 J. The energy stored in the bottom capacitor is U_bottom = ½(20 μF)(4 V)² = 160 J.

(b) Removing the battery changes nothing except that there is no longer an external power supply to keep the three capacitors at the same potential. Removing the dielectric changes the capacitance of the dielectric back to 10 μF. If the 80 μC stayed on the bottom capacitor, its new potential difference would be V_bottom = Q/C = (80 μC)/(10 μF) = 10 V. However the capacitors are in parallel they must always have the same voltage. Thus the charge will move until this occurs. Since the capacitors are now identical, the charge will drop until each capacitor has exactly the same charge. So the extra 40 μC on the bottom capacitor splits three equal ways and each capacitor will end up with a total charge of 160/3 μC. The voltage drop over each capacitor is then V = Q/C = 16/3 Volts. The energy stored in each capacitor with then be U = ½Q²/C = 142.22 J.

(c) The final total energy of the three capacitors is U_final = 3 × 142.22 J = 426.67 J. The initial energy was U_initial = 80 J + 80 J + 160 J = 320 J. The capacitors gained an energy U = U_final - U_initial = 107 J. The extra energy in the system came from the person who pulled the dielectric out. The person had to do work to remove the dielectric since it had a surface charge opposite to the capacitor plates and thus was attracted to it.