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Capacitors, electric field

  1. Dec 7, 2013 #1

    My friend was reading up on transistors and basically the way i see it, a transistor is a device which (using little current/voltage , depending on the transistor) can control larger amounts of current/voltage , so all transistors are devices which work by letting current flow through them or blocking current flow.
    A capacitor has no current flow though it physically (neglecting some small leaks etc) but in a circuit it acts as a device which causes current flow (charge flowing to and from the pates of a capacitor upon charging up and/or discharging)
    After the current/voltage source is disconnected the capacitor still has it's charge due to the electric field, and the opposite charges attracting each other.

    So here is my question , say I wanted to use the capacitor as a source for current flow , now while I charge the capacitor, charges flow to the plates , there is current in the wires ,as the capacitor has charged the current drops until reaches zero.
    There is charge in the capacitor but I want now for that charge to flow back into the circuit or say to another capacitor but the field in the cap won't let me do that , is there a way to shield the electric field between the plates so that the charges won't feel each others attraction anymore and flow back into circuit?
  2. jcsd
  3. Dec 7, 2013 #2
    But the charge CAN flow back into the circuit. Just take a charged capacitor and connect it to a resistor and it will discharge.
    Maybe this video will answer all your questions about capacitors.
    Last edited by a moderator: Sep 25, 2014
  4. Dec 7, 2013 #3
    A nice analogy for a capacitor is the spring. If you push a spring down it will naturally want to push back up as soon as you remove the force that pushed it down.

    A capacitor with almost no capacitance is like a really stiff spring. You can barely move it. In a capacitor's case you can barely charge it. Now if you have two plates really close together so that charges can have some attraction across the plates then the stiffness is counteracted and you can have a more useful spring.

    Dielectrics used in capacitors actually oppose the electric field.

    [itex] \textbf{E} = \frac{1}{\epsilon} \textbf{D}[/itex]

    Higher permitivity means a weaker field. The term D is called the displacement field and it is a function of charge build up on the plates inthe case of a capacitor.

    [itex] D = \frac{Q}{A} [/itex]

    Q is the charge and A is the area.

    Next, voltage in a capcitor is

    [itex]V=E d[/itex]

    d is the distance between the plates.

    Putting it all together:

    [itex] V = \frac{1}{\epsilon}\frac{Q}{A} d [/itex]

    So you can see that with a weaker field (stronger dielectric) there is less voltage on the capacitor per unit charge. The converse is true. You can put more charge on the capacitor using less voltage if the field is weaker.
    Last edited: Dec 7, 2013
  5. Dec 8, 2013 #4
    Well the capacitor only discharges into the resistor it the resistor is placed in parallel with the capacitor, but I have a capacitor whose one plate is connected to a switch which is then connected to ground , the other plate is connected to an inductor which is in series with the capacitor , the inductor is then connected to +ve.

    Why the capacitor is like a spring ? Because when you stop pushing on a spring it comes back but a charged capacitor retains it's charge for a pretty long period of time , the bigger the capacitance the longer,
    And how can a dielectric oppose an electric field ? the dielectric serves to keep the plates electrically separated so that the charges wouldn't run into oneanother but keep at a tiny distance ,
    I will watch the video , but my though was , would it be possible to say charge up the cap , and then somehow turn the dielectric between the plates into a shield so that opposite charges on both plates wouldn't feel each other anymore and run back into the circuit ?

    Maybe I'm getting this wrong, also can a capacitor get discharged through one plate only?
  6. Dec 8, 2013 #5


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    Staff: Mentor

    That's because your applying a voltage to the capacitor. Remove that voltage and the capacitor will discharge back into the circuit.

    Charging a capacitor is like compressing a spring. In order for the spring to then expand you must remove the force compressing the spring. This is analogous to a capacitor in that you charge it using your voltage source and must remove that voltage in order for the capacitor to discharge.
  7. Dec 8, 2013 #6
    Yes drakkith , I understand that in order for the capacitor to discharge it has to be removed from the energy source that charged it.
    What I meant was , yes the capacitor discharges but it does so slowly as compared to a spring which gets back fast to it's original position.

    I watched the MIT lecture that dr.Zoidberg linked to , and I understood most of it but some things I didn't understand, like when you have a capacitor charged to 100 volts, for example
    and you then increase the distance between the plates the potential goes up, ok I see that work is being put into the system so it must show up as something , so it shows up as increased voltage in this case , like if I would have a plate capacitor with 100 volts on it
    and then double the distance between the plates I would then measure 200 volts between the plates ?

    And my last question to which I kinda haven't received an answer is about the shielding of the electric field.
    Usually the way to set up the field or increase /decrease it is by working with charges , as charges have the field around them.Like charging up a capacitor creates a field between the plates and the flow of charge creates current.
    But is it possible to affect the charges themselves and the flow of charges by interacting with the field somehow , like if we could shield the plates of a charged capacitor , the charge would immediately run back into the circuit that is attached to the cap and balance out is that correct ?
  8. Dec 8, 2013 #7
    [itex] V = \frac{1}{\epsilon}\frac{Q}{A} d [/itex]

    What happens to that equation as you increase the distance but hold everything else constant?

    That equation can really answer many of your questions.

    You said something earlier that is incorrent. The dielectric is not there just to block charges from moving across. It supresses the field.

    [itex] E = \frac{1}{\epsilon}\frac{Q}{A} [/itex]

    As the permittivity goes up (stronger dielectric) the field becomes less.

    You can make a capacitor with an air gap. It can even be a vacuum. It won't take much voltage to cause the electrons to jump across but before that limit is reached it will be just like any other capacitor.

    I'm sorry if the spring analogy didn't really do it for you but it's pretty standard stuff. Actually, you can build analog computers that simulate physical systems using that analogy. Any time you have something that acts like a spring you can use a capacitor to simulate it.
  9. Dec 8, 2013 #8
    You can increase the distance between the plates. Then the charges wouldn't "feel" each other as much as before and be pushed back into the circuit more strongly i.e. the voltage would increase.
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