Physics E1bx: Assignment for Feb 24 - Mar. 3 - Canvas ...

Physics E1bx Feb. 24, 2015-Mar. 3, 2015

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Physics E1bx: Assignment for Feb 24 - Mar. 3 Homework #3 Capacitors, Energy, Power, Kirchhoff’s Laws

Due Tuesday, Mar. 3, at 6:00 PM This assignment must be turned in by 6:00 pm on Tuesday, Mar. 3. Late homework can not be accepted. Please write your answers to these questions on a separate sheet of paper with your name and your section TF’s name written at the top. Turn in your homework to the mailbox marked with your section TF’s name in the row of mailboxes outside of Sci Ctr 108. You are encouraged to work with your classmates on these assignments, but please write the names of all your study group members on your homework. After completing this homework assignment, you should be able to…

• calculate various properties associated with a discharging capacitor • calculate various properties associated with a charging capacitor • calculate and interpret the time constant for a capacitor • be able to use and apply Kirchhoff’s junction and loop rules • be able to reduce a given circuit into its equivalent resistor/capacitor • be able to see the similarities between resistors and capacitors in different configurations (i.e.

series vs. parallel) • be able to calculate the energy stored in a capacitor • know that capacitors in parallel have the same potential difference • know that capacitors in series have the same charge • know that the capacitance of a parallel-plate capacitor is only dependent on its geometry


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Here are summaries of this module’s important concepts to help you complete this homework:


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1. Putting Everything Together (Exam-Type Question) (4 pts). Two large flat plates are separated by a distance d. The plates are connected to a battery. The surface area of each plate is A1. (see Figure 2a) A new capacitor is formed by adding two uncharged metal plates, each with area A2, to the capacitor as shown. The battery remains connected (see Figure 2b).

a) Write an expression for the surface charge density on the plates

of the capacitor. Is the charge density greater than, less than, or equal to the charge density on the plates before the new plates were attached? Explain.

Recall that surface charge density is defined as the ratio of charge and area:

𝜎 =𝑄𝐴

Now, capacitance is defined as the ratio between charge and potential difference.

𝐶 =𝑄𝑉

We can substitute 𝑄 = 𝜎𝐴 and 𝐶 = 𝜀!

!! into the above expression and simplify:

𝜀!𝐴𝑑 =

𝜎𝐴𝑉

𝜀!1𝑑 =

𝜎𝑉

𝝈 = 𝜺𝟎𝑽𝒅

As can be seen, area A drops out of this equation so, for a parallel-plate capacitor, surface charge density only depends on the fundamental constant 𝜀!, the potential difference V, and the separation distance d of the two plates. Because the apparatus is still connected to the same battery and the separation distance has not changed, the charge density with the new plates attached is equal to the original charge density.

b) Write an expression for the total charge on one of the plates of the capacitor. Is this total charge greater than, less than, or equal to the total charge on one of the original plates? Explain.

d

Area A1

V0

Figure 2a.

Area A1

d

V0

Area A2

Figure 2b.


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The total charge on a plate can be written in terms of surface charge density σ and area A.

𝑄 = 𝜎𝐴 With the new plates added, the total area is

𝐴 = 𝐴! + 𝐴! So, the total charge on one of the plates is

𝑸 = 𝝈 ∗ 𝑨𝟏 + 𝑨𝟐 We found in the previous part that surface charge density is a constant. Thus, because the area increased, the amount of charge on a plate also increased.

c) Use the definition of capacitance to find the capacitance of the pair of plates. Has the capacitance increased, decreased, or remained the same?

The definition of capacitance is

𝐶 = 𝑄𝑉

We can substitute in our results from the previous two parts to obtain an expression for the capacitance of the new parallel-plate capacitor:

𝐶 = 𝜎 ∗ 𝐴! + 𝐴!

𝑉

𝐶 = 𝜀!𝑉𝑑

𝐴! + 𝐴!𝑉

𝑪 = 𝜺𝟎𝑨𝟏 + 𝑨𝟐

𝒅 Compared with the result from part a), we see that capacitance has increased. d.) Now consider two capacitors connected in parallel to the same battery, one of area A1 and the other of area A2. Which capacitor has more charge on it (assume A1> A2)? They have the same charge density; that means the plate of area A1 has more charge on it since it is larger.


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2. Energy in a discharging capacitor (2 pts). Consider a capacitor C with an initial charge q0. At time t = 0, the switch S is closed and the capacitor begins to discharge through resistor R. a) How does the total energy dissipated in the resistor relate to the total

energy initially stored in the capacitor? Explain. The system (i.e., the circuit) only contains the resistor and the capacitor. With energy conservation in mind, all of the initial stored electrical potential energy must be converted to heat through ohmic heating in the resistor. b) Derive an expression for the energy stored in the capacitor as a function of time. Express your

answer in terms of q0, C, R, and t. The energy stored in the capacitor can be expressed as:

U =12Q2

C.

For a discharging capacitor, we know that the charge as a function of time can be written

Q t( ) = q0e−tRC

so we can just plug that into the first expression to get U(t):

U t( ) = 12q02

Ce−2tRC .

3. Capacitors (Exam-Type Question, 1 pt). Four uncharged capacitors with equal capacitances are combined in parallel. The combination is connected to a 6.09-V battery, which charges the capacitors. The charging process involves 2.37 × 10-4 C of charge moving through the battery. Find the capacitance of each capacitor.


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4. Resistors and Capacitors (Exam-Type Question, 2 pts). Consider the circuit shown in the diagram below. Before the switch is closed, both capacitors are uncharged.

a.) Immediately after the switch is closed, what is the amount of current supplied by the battery?


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b.) Assuming the switch remains closed for a very long time, which capacitor will be the first to reach 95% of its final charge level? What is the time constant for charging this capacitor?

5. Splitting up the Voltage (Exam-Type Question, 1 pt) Four resistors are connected to a battery as shown in the figure. The current in the battery is I, the battery emf is ℇ = 6.15 V, and the resistor values are R1 =


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R, R2 = 2R, R3 = 4R, R4 = 3R. Find the voltages across each resistor.

NOTE: the RESISTOR VALUES ARE DIFFERENT than in your assignment, but the approach is the same.


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The numbers you should have gotten: Through R1: 2.05 V Through R2: 1.37 V Through R3: 2.73 V Through R4: 4.10 V

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