Circuits I

  1. A capacitor consists of two _______ or ______ of any shape. The conductors have charges of equal _______ and opposite ______ (__), and a _________ ________ (__) between them. Net charge on a capacitor is ______. When we talk about the charge on a capacitor, we always mean the magnitude of charge on either plate which is + Q. We say that the capacitor stores ______.
    • conductors or electrodes 
    • equal magnitude 
    • opposite sign (Q)
    • potential difference (ΔV)
    • zero
    • charge
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  2. In a capacitor, the electric field strength E and the potential difference ΔVC _______ as the charge on each electrode ________. Define the capacitance, C and state the identity
    • increases 
    • increases
    • Capacitance, C: The capacitance of a capacitor is the ratio of the magnitude of the charge on either conductor to the potential difference between the conductors 
    • C Ξ Q/ΔV > 0
  3. In a capacitor, when a SMALL amount of +charge is moved from one _____ to the other, the electric field and potential difference ΔV are _____. When MORE charge is moved from one ______ to another, the electric field and potential difference _______
    • electrode or conductor
    • small
    • electrode or conductor 
    • increase
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  4. Capacitance is always _______. The SI unit of capacitance is a ______ (__). Typically you will see units of ________ (__) and _______ (__).
    The capacitance of a device depends on the ________ arrangement of the conductors: 
    ♜______, _____, and _______ of the two electrodes
    ♜ A capacitor with a ______ ______ holds more charge for a given potential difference than one with _______ _______
    • always positive
    • farad (F)
    • microfarads (μF) and picofarads (pF)
    • geometric 
    • Shape, size, and spacing
    • large capacitance
    • small capacitance
  5. The field is _______ in the central region between the plates, and is _______ at the edges of the plates
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    As long as the separation between the plates is _____ compared with the dimensions of plates, the edge effects can be ignored
    • uniform
    • nonuniform 
    • small
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    • C = Q/ΔV
    • C = (4 nC)/(2.0 V)
    • C = 2 nF
    • C. 2 nF
  7. What must take place in order to charge a capacitor? The simplest way to do this is to use a source of _______ ________ such as a battery. A battery uses its internal chemistry to maintain a fixed _______ ________ between its ________
    • To charge a capacitor, we must move charge from one electrode to the other
    • potential difference 
    • potential difference
    • terminals
  8. Charge flows from the top electrode leaving it _______. The charge then flows through the ______ which acts as a _______ ______. The charge ends up on the bottom electrode, making it _______ charged. 
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    The movement of the charge stops when ΔVC is equal to the ______ ______. The capacitor is then ______ ______
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    If the battery is removed the capacitor _______ ________, with ΔVC equal to the _______ ________
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    • negative
    • battery
    • charge pump
    • positively 
    • battery voltage
    • fully charged
    • remains charged
    • battery voltage
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  9. Two parallel plates of equal area (A) separated by a distance (d) (assume vacuum in between the 2 plates)
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    d << dimensions of the plates, so ______ effects can be ignored. Magnitude of the charge per unit area on either plate is: 
    σ = ____
    Electric field is ______ between the plates and _____ elsewhere. 
    E = _____ = _____ and ΔV = _____ = _____
    C = ____ = _____
    • edges 
    • σ = Q/A
    • uniform 
    • zero
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    **σ is just lower case sigma (Σ upper case)
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    • Something (probably a battery) was used to generate charge separation (Q) between the electrodes. Then it was a removed, and we know when it (the battery) is removed the capacitor remains charged, so Q is constant 
    • ΔV = Ed = Qd/ε0A
    • E. Both remain constant
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    • ΔV = Ed if d↑, then E↓ (so B, C and E eliminated) 
    • C = ε0A/d if C↓ (because d↑) then Q↓ since ΔV is constant (A is eliminated)
    • So D.
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    • A = 1 m2 
    • d = 1*10-3 m
    • a) C = Q/ΔV = ε0A/d
    • C = (8.85*10-13 F)(1)/(1*10-3 m)
    • C = 8.85*10-9 F
    • b) Q = CΔV & ΔV = 100 V
    • Q = (8.85*10-9 F)(100 V) = 8.85*10-7 C
  13. Define dielectric and give 3 examples
    If the dielectric completely fills the space between the plates, the capacitance ________ by the dimensionless factor κ, called the ______ ______ of the material:
    Cwith dielectric = ________
    • A Dielectric: an insulating material that increases capacitance when placed between the plates of a capacitor
    • Waxed paper, rubber, plastic
    • increases
    • dielectric constant
    • Cwith dielectric = κCwithout delectric
  14. Polar molecules are _______ oriented in the absence of an external electric field. When an external electric field is applied, the molecules partially _______ with the field. The charged edges of the dielectric can be modeled as an additional pair of _______ ______ establishing an ______ _____ (___) in the direction opposite that of _____
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  15. Label the diagram
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    Field (arrow right above) due to _______ ______ on dielectric.
    (Blue arrow): The net field is the vector sum of the _______ field and the field due to the _______
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    • induced charges 
    • applied field 
    • dielectric
  16. State 3-story for Case 1
    Then state what happens to:
    C
    Q
    V
    E
    U(check this answer with prof.)
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    • UE decreases by a factor of κ such that (UE with dielectric)= (1/κ)(UE without dielectric)
  17. State 3-story for Case 2
    Then state what happens to:
    C
    Q
    V
    E
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  18. For a parallel-plate capacitor with dielectric:
    C = ______
    In theory, d could be made very small to create a very ______ capacitance. 
    In practice, there is a limit to d (explain). For a given d, the maximum voltage that can be applied to a capacitor without causing a discharge depends on the _______ ______ of the material
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    • large 
    • d is limited by the electric discharge that could occur through the dielectric medium separating the plates 
    • dielectric strength
  19. A dielectric provides the following advantages (3):
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    • a) The battery is removed so Q stays the same
    • Q = CV = (200 pF)(100 V) =  20000 pC (constant)
    • b) decreases by factor of κ = 2.0
  21. The work done in charging the capacitor appears as ______ ______ ______ (___). Or a charged capacitor stores energy as ______ ______ _____: 
    UE = _____ = _____ = _____ = _____
    This applies to a capacitor of any _______
    The energy stored increase as the charge ______ and as the potential difference ______. In practice, there is a ______ voltage before discharge occurs between the plates
    • electric potential energy (U)
    • electric potential energy
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    • geometry
    • increases
    • increases 
    • maximum
  22. A capacitor can charge very slowly and then can release the energy very _______. A medical application of this ability to rapidly deliver energy is the ________. 
    Define Fibrillation
    • quickly
    • defibrillation 
    • Fibrillation: the state in which the heart muscles twitch and cannot pump blood. 
    • **A defibrillator is a large capacitor that can store up to 360 J of energy and release it in 2 milliseconds. The large shock can sometimes stop fibrillation
  23. The energy stored in a capacitor can be modeled as being stored in the electric field between the plates of the capacitor:
    UE = ______ = ______
    The energy per unit volume, called the energy density, is:
    uE = ________ = ______
    The energy density in any electric field is _______ to the square of the magnitude of the electric field at a given point. The energy density has units ______
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    • proportional 
    • J/m3

    • **Note: (Ad) is basically the volume
    • **Note: ue = Ue/volume =energy density and is proportional to E2
    • This expression is for any kind of capacitor (shape doesn't matter)

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    • Ue = 1/2CV2
    • 2 mJ = 1/2C(1.5)2
    • C = (4*10-3)/(1.5)2 = 1.78*10-3 F
    • Energy stored in the capacitor charged to 3.0 V: 
    • Ue = 1/2(1.78*10-3 F)(3 V)2 
    • E. Ue = 8 mJ
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    • Ue = 1/2CV2
    • C = 2Ue/V2 = 2(8.4*106)/(23500)2
    • C = 3.04*10-2F
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Author
chikeokjr
ID
346749
Card Set
Circuits I
Description
Circuits I
Updated