OBJECTIVE: 

1) I will be able to derive an expression for the capacitance of a parallel plate capacitor after today's class

2) I will be able to calculate the energy stored in a capacitor after today's class

3) I will be able to derive an expression for the energy density in a capacitor 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
• U_E= Q²/2C = 1/2Q∆V = 1/2 C(∆V)² (energy stored in a capacitor)

CALENDAR:

Quiz on Thursday

WORK O' THE DAY -

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I need to emphasize something from the chapter yesterday: How to evaluate charge on a capacitor:

The total charge of the system is the same as the charge on each cap in series.. That means no matter how many capacitors you have in series, the charge remains the same across each one.

 

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ENERGY OF A PARALLEL PLATE CAPACITOR

Let's again briefly review a capacitor present in an electronic circuit:

Notice in this situation that the chemical energy stored in the battery is transformed into electrical potential energy stored in the capacitor.

If we graph the change in electrical potential vs charge stored in the plates of a capacitor we get a straight line:

Notice also that the slope of the line is also of interest:

1/C

Through just a bit of integration (found on page 787 but I'm not going to require that you memorize or practice that) we can find that the Work done to move a charge through that potential difference is:

Q²/2C

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Recall the relationship between electric potential energy, charge and electric potential:

U = Q∆V

From this relationship we can find the area under the 'curve'. The 'curve' here is a triangle so the area of that triangle is given as 1/2Bh or:

U_E=1/2Q∆V

 

And since UE = W we can also add from above:

U_E= Q²/2C = 1/2Q∆V = 1/2 C(∆V)²

(I dunno about you, but methinks that is DEFINITELY FLASH CARD TIME!)

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Please work through Example 26.3 (I forgot to have you do that yesterday) and 26.4

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HOMEWORK:

HW Problems Conceptual 26.7 (yes, we know that capacitors do store enery, but HOW and WHERE is it stored)?, 8 on page 800 and problems P26.30, 32, 35 on pages 802 and 803

 

 

ANSWERS: