OBJECTIVE:  I will be able to calculate the capacitance between two charged plate capacitors after today's class.

WORDS/FORMULAE FOR TODAY

TERMS:

• Capacitor - two charged surfaces that can store electrical energy

CONSTANTS:

• k_e = 8.987 x 10⁹ Nm²/C²
• k_e = 1/4πε_o

UNITS:

• Capacitance = C
  • (SI Units "farads" = F)
  • C = Q/∆V
  • farads are always positive
  • capacitance measures ability of the system to store charge
  • 1 farad is a MASSIVE amount of storage so we will typically talk in microfarads (μF = 10^-6 F) or picofarads (pF = 10^{-12 F})

FORMULAE:

• Capacitance:
  • C = ε_oA/d (for a parallel plate capacitor)
  • C = Q/∆V (generally)
• Electrical Field: For a parallel plate capacitor: E= σ/ε_o

WORK O' THE DAY: 

TEST RETURNS FIRST

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WORK O' THE DAY - Capacitors

Recall from the beginning of the term that a thundercloud results in a build up a negative charge at the bottom of the clouds. That negative charge then induces a positive charge in the ground.

Lets consider a similar situation where instead of allowing charge to build up we force a potential difference between two surfaces.

Consider the case of two plates of conducting material placed a distance 'd' apart from one another. If we connect a wire to the (-) terminal in a battery to one plate that was previously uncharged, an electric potential would form between the battery and the plate. Electrons would then flow from the battery to the plate resulting in the plate acquiring a uniform negative charge.

That flow would continue until the potential of the battery and the potential of the plate were equal.

By similar construction, if the other plate were connected to the positive terminal of the battery, electrons would flow *from* the plate *to* the battery resulting in a build up of positive charge on that plate.

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Capacitors - general notes:

• Capacitance is the ability for two charged surfaces to store electrical energy (hence the term 'capacity').
• SI Units are 'farads': f
• Capacitance is defined as the total charge divided by the change in potential between two surfaces: C = Q/∆V
• 1 farad is a massive amount of stored energy. We will typically be working with microfarads (10^-6) or picofarads (10^-9)

 

We'll be talking about 3 physical types of capacitors determined by shape:

1) Parallel Plate capacitors -

• Parallel plate capacitors are characterized by two plates of equal potential separated by some distance d
• The surface area and therefore the surface area charge density (σ) are very important: Recall that in our previous lesson we found that the electric field between two charged plates is found to be uniform: E= σ/ε_o
• The good news is the math for parallel plate capacitors is pretty easy--- plugging and chugging sorts of stuff

2) Spherical capacitors -

• Spherical capacitors are characterized by two spherical surfaces of equal potential separated by some distance d
• The GOOD NEWS is we are NOT nearly as interested in surface area charge density for spherical capacitors
• The BAD NEWS(?) is that means lots of integration

3) Cylindrical capacitors -

• Cylindrical capacitors are characterized by two cylindrical surfaces of equal potential separated by some distance d
• The GOOD NEWS is we are NOT nearly as interested in surface area charge density for spherical capacitors
• The BAD NEWS(?) is that means lots of integration

 

Parallel Plate Capacitors in Action:

the plates then are a source of electric potential energy and in a very real sense actually store energy. Unfortunately this storage mechanism doesn't hold energy very efficiently and energy is lost at a predictable rate (we'll talk about that later).

Let's take a look at a life-saving instance of a capacitor in action.... a defibrillator.

During one particular type of heart attack, the heart muscle ceases its usual 'sinusoidal' rhythms that ensure an appropriate amount of oxygenated blood to the rest of the body.

An early electric engineer discovered that applying a significant 'shock' to the heart muscle could actually reset the regular contractions of the heart and return it to its regular rhythm:

Here's a description of what follows (courtesy of the 'AnasthesiaUK' web site): http://www.frca.co.uk/article.aspx?articleid=100392

Figure 2 shows a defibrillator.

When the switch is in position 1, direct current (DC) from the power supply is applied to the capacitor. Electrons flow from the upper plate to the positive terminal of the power supply and from the negative terminal of the power supply to the lower plate.

Therefore current flows and a charge begins to build up on each electrode of the capacitor, with the lower plate becoming increasingly negatively charged, and the upper plate increasingly positively charged.

As the charge builds up on the plates, it creates a potential difference across the plates (V), which opposes the electromagnetic force of the power supply (Emf). Initially when there is no charge on the plates, V is zero and it is easy to move electrons onto the plates. As V increases, however, it opposes further movement of electrons, and increasing work must be done to move more electrons onto the plates.

The work done (W) to move charge (Q) through a potential difference V is: W = VQ. Charging a capacitor is therefore an exponential process, with a time constant determined by the capacitance and the resistance of the circuit through which the current flows (Figure 3). When V equals E, the current ceases to flow and the capacitor is fully charged. In this example, the amount of charge stored (Q = CV) is 32 µF x 5000 V = 160 mC.

You may be interested to know that defibrillators can actually be surgically implanted as was the case with a former U of W woman's basketball star Kayla Burt.

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You folks who are interested in the (*gasp*) life sciences, may be interested to know that defibrillators DO NOT 'shock' a flatlined heart back into normal rhythm.... let's take a quick gander at THIS

From what you know of a capacitors so far, suggest a reason why capacitors are used to 'jump start' the heart muscle instead of simply applying live wires to either side of a patient's chest.

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If time permits:

Let's quickly review a simple form of integration:

1. We start with the basic knowledge that the circumference of a circle = 2πr
2. at some infinitely small point, we'll call it dr, the circumference of that tiny dot = the area of that tiny dot.
3. If we increase that tiny bit by that same tiny increment dr we have a slightly bigger area
4. We can basically add up all of those circumferences (circumference?):

• ∑_0^i=R 2πr which shakes out into the integral ∫2πrdr (with the same limits)
• integrating from i = 0 to i = R we get 1/2(2πr²) =πR² (where R is the numeric value for r)

this is sometimes called the 'shell method' or the 'onion method'. In other words we are summing up all the 'shells' or 'skins' to find the sum of all them together = the area within

HOMEWORK:

1) Write a reflection on this test, turn it in on Wednesday

2) Take a look at Problem 26.4 (and 26.5 if time permits, in fact, try integrating 26.5 in a DIFFERENT way)

Use integration to find ∆V and then worry about Q later on. The goal here is to practice using integration in a meaningful way....in this case to find ∆V

Be prepared to EXPLAIN how to do that entire process from soup-to-nuts

That means:

• Show us how to start with: v_b - v_a = -∫E ∙ ds and end up with the capacitance of the cylinder

• Do not skip steps like the book does

• Be prepared to explain EXACTLY what is happening during the integration process <HINT: Do this the way that YOU think makes the most sense... following the steps the book presents will almost certainly NOT do that. MAKE IT MAKE SENSE PEOPLE.

  • That means annotations -- just like when you are writing a lab