Week 4 - Capacitors and Capacitance

Sheldon, you don’t give your mother enough credit. She’s warm, she’s loving, she doesn’t glue electrodes to your head to measure your brain waves while potty training. Leonard Hofstadter in “The Big Bang Theory” s. 2 ep. 15

Exercise 4.1: Capacitors A parallel-plate capacitor is placed in vacuum with a separation of 10.0 mm and an area of 1.00 m2 . It is connected to a potential difference of 20.0 kV. a) What is the capacitance? Answer: C = ε0

A = 0.885 nF d

(1)

b) What is the charge on each plate? Answer: Q = CV = 17.5 µC

1

(2)

c) What is the magnitude of the electric field? Answer: E=

σ Q = = 2.00 × 106 N/C ε0 ε0 A

(3)

d) What is the energy stored on the capacitor? Answer: U=

Q2 (17.5 µC)2 = = 0.173 J 2C 2 · 0.885 nF

(4)

Exercise 4.2: Discussion Questions a) A parallel-plate capacitor is charged by being connected to a battery and is then disconnected from the battery. The separation between the plates is then doubled. How does the electric field change? The potential difference? The total energy? Explain your reasoning. Answer: The electric field is constant. The potential difference increases. The total energy increases. (This can be easily seen from the different relations between capacitance, distance, voltage, etc. b) The freshness of a fish can be measured by placing a fish between a capacitor and measuring the capacitance. How does this work? Answer: With time the moisture of the fish decays, causing the dielectric constant to change. If we know how fast the fish loses its moisture we can make a relation between the capacitance and the freshness. c) The charged plates of a parallel capacitor attract eachother, so if you pull the plates a distance ∆d apart you have to do work. How much work is required? Answer: Equal to the new energy stored. W = ∆U =

Q2 2∆C

=

Q2 ∆d 2ε0 A

Exercise 4.3: Camera flash capacitor Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for 1/675 s with an average light power output of 0.27 MW. a) If the conversion of electrical energy to light is 95% efficient (the rest of the energy goes into thermal energy), how much energy must be stored in the capacitor before the flash? Solution:

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Let the energy in the flash be Ef and the energy stored in the capacitor be E. Then we have the relation   1 1 1 0.27 MW × Ef = 0.95E ⇒ E = Ef = s = 421 J (5) 0.95 0.95 675 where we have used the fact that 1 W = 1 J/s.

Answer: E = 421 J

(6)

b) The capacitor has a potential difference between it’s plates of 125 V when the stored energy equals the value calculated in part (a). What is the capacitance? Solution: Using the formula for energy stored in a capacitor which relates capacitance, potential and energy, E = 12 CV 2 , we get a capacitance of

C=

2 × 421 J 2E = 2 = 54 mF. V2 (125 V)

(7)

Answer:

C = 54 mF

(8)

Exercise 4.4: How is a computer keystroke registered?

d A Figure 1: A keyboard application of a plate capacitor.

One application of a capacitor is to use it as a sensor. This is used in some computers. Consider the button in figure 1 with A = 50 mm2 and d = 0.6 mm. When you push a button, the distance Week 4 – September 10, 2012

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d decreases and the capacitance goes up. When the capacitance get bigger than C 0 = 1.250 pF the keystroke is registered. Find the distance we need to press the button down in order for the keystroke to be registered. Answer: d≤

Aε0 C0

Solution: So the keystroke is registered when the capacitance is bigger than some minimal capacitance C 0 . Thus C=

Aε0 Aε0 Aε0 ≥ C0 ⇔ d ≤ 0 ≤ 0 d Cd C

Exercise 4.5: Combining Capacitors Several 0.25 µF capacitors available. The voltage across each is not to exceed 600 V. You need to make a capacitor with capacitance 0.25 µF to be connected across a potential difference of 960 V supplied by a battery. a) Show in a diagram how an equivalent capacitor with the desired properties can be obtained. Solution: Since we can not expose the capacitors to 960 V we have to put some capacitors in series to decrease the voltage across each capacitor. Then since the capacitors are equal we will get half the voltage 480 V across each of them. But the capacitance will be lower. It will be

 C=

1 1 + C1 C2

−1

 =

1 1 + 0.25 µF 0.25 µF

−1 = 0.125 µF

(9)

which is half of the capacitance we really want. But if we combine two such capacitors in parallel the capacitance will double! A sketch of the resulting combination of capacitors is shown in figure 2

Answer: b) Suppose that you ignored the advice of not letting the voltage across each of the capacitors exceed 600 V. What would happen when you connected the the battery of 960 V across the capacitor? Solution: See ’Dielectric Breakdown’ on page 833 in the book.

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Figure 2

Exercise 4.6: Cylindrical capacitor as fuel sensor Capacitors can be used as a fuel sensor, keeping track of the fuel levels in vehicles. These type of fuel level sensors are commonly used in large transport aircraft. Let’s study these sensors in some detail. Consider the coaxial conducting cylinders in figure 3 of radii a and b and length L.

L

b a

Figure 3: Capacitor consisting of two coaxial cylinders.

a) Given that the inner cylinder has charge −Q and the outer has Q. Ignore the fringing fields and use Gauss law to find the electric field between the cylinders expressed in terms of the cylinders surface charge σ. Solution: Construct an imaginary cylinder of length L in between the two cylinders. Because the positive charge is on the outer cylinder the field will anti parallel to dA at every point which gives I E · dA = −E2πLr =

⇒E=−

Q ε0

σa Q ˆ r=− ˆ r 2πε0 Lr ε0 r

where the surface charge is given by σ = Q/2πaL.

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Answer: E=−

Q ˆ r 2πε0 Lr

b) Show that the capacitance of the arrangement is given by 2πε0 L ln ab

C=

Solution: The capacitance is defined as C = Q/V so we first have to find the potential difference between the two cylinders V . Z

b

V = Vb − Va = − a

Q E · dr = 2πε0 L

Z

1 Q b dr = ln r 2πε0 L a

we then get C=

Q Q 2πε0 L

ln ab

=

2πε0 L . ln ab

c) Find the energy stored in the electric field between the cylinders given that a = 1 mm, b = 2 mm and Q = 0.2 nC. Solution: P In general the energy between different objects i with charge Qi can be written U = 12 i Qi Vi . For a capacitor with opposite charge on the two conductors this implies that U = 12 Q∆V , and C = Q/V so that

U=

1 Q2 2 C

U=

1 Q2 2 C

Answer:

d) Suppose now that we place this capacitor partially in water with K = 80 and that the total charge on the capacitor is unaffected. See figure 4. Find an expression for V in the two regions. Hint: The polarized water will change the surface charge on the part of the capacitor immersed Answer: Vair =

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σair a b ln ε0 a

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a

b

z

Figure 4: Cylinder capacitor used as a water height sensor.

Vwater =

σwater a b ln Kε0 a

.

Solution: Eair = −

σair a ε0 r

while Ewater = −

σwater a Kε0 r

which gives Vair =

σair a b ln ε0 a

and Vwater =

σwater a b ln Kε0 a

. e) Now we know that a conductor is an equipotential. Can you use this to find a relation between the surface charge in the two regions? Use this to find the capacitance C = C(z). Answer: σwater = Kσair

C=

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Q (z(K − 1) + L) = 2πε0 . V ln ab

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Solution: A conductor is an equipotential Vair = Vwater which means that σwater = Kσair . Now the total charge on the capacitor is Q = σwater 2πaz + σair 2πa(L − z) = 2πaσ (z(K − 1) + L) such that C=

Q (z(K − 1) + L) . = 2πε0 V ln ab

You have created a fluid level sensor!

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