Practice Problems: Capacitors Solutions

1. (easy) Determine the amount of charge stored on either plate of a capacitor (4x10^-6 F) when connected across a 12 volt battery.
C = Q/V
4x10^-6 = Q/12
Q = 48x10^-6C

2. (easy) If the plate separation for a capacitor is 2.0x10^-3 m, determine the area of the plates if the capacitance is exactly 1 F.
C = ε_oA/d
1 = (8.85x10^-12)A/(2.0x10^-3)
A = 2.3x10⁸ m²

3. (moderate) Calculate the voltage of a battery connected to a parallel plate capacitor with a plate area of 2.0 cm² and a plate separation of 2 mm if the charge stored on the plates is 4.0pC.
Area = 2.0 cm²(1 m/100cm)² = 2.0x10^-4 m²
C = ε_oA/d
C = (8.85x10^-12)(2.0x10^-4)/(2.0x10^-3)
C = 8.85x10^-13
C = Q/V
8.85x10^-13 = 4.0x10^-12/V
V = 4.5 volts

4. (easy) A parallel plate capacitor is constructed of metal plates, each with an area of 0.2 m². The capacitance is 7.9nF. Determine the plate separation distance.
C = ε_oA/d
7.9x10^-9 = 8.85x10^-12(0.2)/d
d = 2.2x10^-4 m = 0.22 mm

5. (easy) A capacitor (parallel plate) is charged with a battery of constant voltage. Once the capacitor reaches maximum charge, the battery is removed from the circuit. Describe any changes that may take place in the quantities listed here if the plates were pushed closer together.
a. Charge (The charge deposited on the plates doesn't change when the battery is removed and thus the charge and the charge density remains the same as the plates are moved closer together.)
b. Capacitance (Since the capacitance is C = ε_oA/d, and the area is not changing, any decrease in plate sepration (d) will cause an increase in capacitance.)   
c. Voltage (Because C = Q/V, and the charge doesn't change, an increase in capacitance implies a decrease on voltage.)
d. E-field (Since ΔV = -Ed, the E-field will remain the same as both the voltage and the distance decrease proportionately.)
e. Would any of the answers above change if the battery was not disconnected from the battery? Explain your response. Yes. The voltage would not change if the battery remained connected to the capacitor. The capacitance would still increase because it is based solely on the geometry of the capacitor (C = ε_oA/d). The charge would increase because Q = CV and the capacitance increased while the voltage remained the same. Finally, the E-field would increase because E =  |ΔV/d| and the distance across the plates decreased while the voltage stayed the same.

6. (moderate) Random access memory chips are used in computers to store binary information in the form of "ones" and "zeros". One common way to store a "one" is to charge a very small capacitor. Of course, the same capacitor without charge represents a "zero". A memory chip contains millions of such capacitors, each coupled with a transistor (that acts as a switch), to form a "memory cell". A typical capacitor in a memory cell may have a capacitance of 3x10^-14 F. If the voltage across the capacitor reading a "one" is 0.5 v, determine the number of electrons that must move on the the capacitor to charge it.
C = Q/V
3x10^-14 = Q/(0.5)
Q = 1.5x10^-14 C
#electrons = Total charge/Charge per electron
#electrons = 1.5x10^-14/1.6x10^-19
#electrons = 93750 electrons

7. (easy) C₁ = 10 F and C₂ = 5 F. Determine the effective capacitance for C₁ and C₂ connected in series and in parallel.
In series:
1/C = 1/C₁ + 1/C₂
1/C = 1/10 + 1/5
C = 3.3 F
In parallel:
C = C₁ + C₂
C = 10 + 5 = 15 F

8. (moderate) If the two capacitors in question #7 were connected to a 50 volt battery determine the voltage across the capacitors for each connection type.
For the series connection:
The charge on each capacitor is the same as the charge on the effective capacitance.
C = Q/V
3.3 = Q/50
Q = 165 C
For the 10F capacitor:
10 = 165/V
V = 17 volts
For the 5 F capacitor:
5 = 165/V
V = 33 volts
For the parallel connection:
The voltage is the same (50 v) across each capacitor.

9. (moderate) Evaluate the circuit shown below to determine the effective capacitanceand then the charge and voltage across each capacitor.

The equivalent capacitance is 4 μF. The voltage across the equivalent capacitor is 20 volts.
This voltage is also across both of the 2 μF capacitors that were created by the series combinations in each branch.
Find the charge on each 2 μF capacitor:
C = Q/V
2 μF = Q/20
Q = 40 μC
The 4 μF capacitors in each branch have the same charge as the 2 μF capacitors. Use this to find the voltage across each:
C = Q/V
4 μF = 40 μC/V
V = 10 volts
In summary, each of the original 4 μF capacitors have a charge of 40 μC and a voltage of 10 volts.

10. (moderate) Evaluate the circuit shown below to determine the effective capacitance and then the charge and voltage across each capacitor.

The effective capacitance is 6 μF with a voltage of 100 v.
The voltage across the 4 μF and the 2 μF capacitors is also 100 v
The charge on the 4 μF capacitor:
C = Q/V
4 μF = Q/100
Q = 400 μC
The charge across the 2 μF capacitor:
C = Q/V
2 μF = Q/100
Q = 200 μC
All three 6 μF capacitors also have 200 μC of charge.
Find voltage for the 6 μF capacitors:
C = Q/V
6 μF = 200 μC/V
V = 33.3 v

11. (moderate) Evaluate the circuit shown below to determine the effective capacitance and then the charge and voltage across each capacitor.

The equivalent capacitance is 6 μF. The voltage across the equivalent capacitance is 40 v as is the voltage across the 3 μF capacitors and is the same as the 1 μF and 2 μF capacitors.
Find the charge on the 1 μF capacitor:
C = Q/V
1 μF = Q/40
Q = 40 μC
Find the charge on the 2 μF capacitor:
C = Q/V
2 μF = Q/40
Q = 80 μC
Find the charge on the 3 μF capacitors:
C = Q/V
3 μF = Q/40
Q = 120 μC
This is the same charge on each of the 6 μF capacitors.
Find the voltage on each of the 6 μF capacitors:
C = Q/V
6 μF = 120 μC/V
V = 20 v