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Homework Help: MicroF capacitor problem

  1. Dec 17, 2004 #1
    A i microF capacitor is connected to a 12 V battery, charged up, and then disconnected to the battery. Afterwards a dielectric, (with dielectric constant = 2), is inserted between the plates of the capacitor. What is the energy stored in the capacitor and discuss where the energy has gone or come from. WHat would happen if you then reconnected the capacitor to the battery.

    I can't figure this one out. Please help me as quickly as possible. Thanks.
  2. jcsd
  3. Dec 17, 2004 #2

    Andrew Mason

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    Find the potential energy of the capacitor before insertion of the dielectric. This will be in terms of the voltage between the plates and its capacitance. (As you probably know, a capacitor will collect charge until the voltage between the plates is equal to the applied voltage (12 v.), so you know the voltage between the plates) .

    Since capacitance is directly proportional to the dielectric constant (space = 1; air [itex] \approx[/itex] 1), when the dielectric constant is doubled, the capacitance doubles. (this is because the material with the higher dielectric constant provides an opposing electric field between the charges which reduces the field between the plates).

    C = Q/V and Q remains the same since no charge leaks from the capacitor with the circuit disconnected. So if C doubles, V has to reduce by a factor 1/2. That should tell you what happens when you reconnect the 12.v. battery.

    Work out the potential or stored energy in the capacitor after insertion of the dielectric. There will be a reduction due to the reduced field. Where did it go? It must have gone into building the opposing electric field in the dielectric, essentially by aligning polar molecules in the dielectric material with the electric field between the plates. [Edit: As the dielectric is inserted the alignment of the molecules in the dielectric will create an attractive force between the charge on the plates and the dielectric. This will pull the dielectric forward so electric potential energy is converted into mechanical energy of the dielectric as well as energy in aligning the molecules].

    Last edited: Dec 18, 2004
  4. Dec 17, 2004 #3


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    Well, what happens when a voltage is applied to a capacitor? Hint: EMF (1 volt = 1 J/C)

    Capacitors and inductors are storage devices. What is the function for Energy stored in the capacity and it's capacitance? How does a capacitor store energy?

    What is the significance of a dielectric? What affect would it have?

    What happens when the battery is reconnected? Well if it is connected with the same polarity - the voltage is reapplied. What happens.

    If the polarity is reversed - what happens? [Don't try that by the way.]
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