capacitors and dielectrics

Mark Kinsler

We generally do not normally work in units of electrical charge, so the concept seems a bit strange.  The easiest way to deal with this is to use the relationship I=Q/t, which says that current is the rate at which charge is delivered.  So we rewrite this definition as Q=It, which says that we can calculate the charge we've placed on a capacitor's plates by multiplying the current through the capacitor by the time we've applied that current.  Result: to determine charge, you need a watch and an ammeter.

If we charge a capacitor to a voltage V by pushing a charge Q (=It) onto the plates, we will find that the relationship between the voltage and the charge is Q/V=C, which is the definition of capacitance.  Note that we can thus measure the capacitance of a capacitor with an ammeter, a watch, and a voltmeter.

Now watch this:  We've charged the capacitor to voltage V.  We rewrite the definition of capacitance Q/V=C ---> V=Q/C.  This predicts what will happen if we reduce the capacitance by sliding the dielectric out from between the plates.  Yup: the voltage will go up, because we won't be reducing the charge on the plates.  C goes down, V goes up.  You can do the same thing by just pulling the plates apart.

The opposite will happen if we increase the capacitance of a capacitor.  If we were to charge an air dielectric capacitor to a voltage of 700 volts and then slip a piece of glass between the plates such that the glass completely fills the space between the plates, we'd find that the voltage has dropped to 100 volts.  Thus we say that glass has a relative dielectric constant of seven.  The dielectric constant of air and vacuum are equal for most practical purposes.

M Kinsler
512 E Mulberry St. Lancaster, Ohio USA 740 687 6368

_________________________________________________________________
Get your FREE download of MSN Explorer at

Doug Hale

#1019

Mark,
  I looked at your website - a nobel cause.

   Are you familliar with Marshal Brain    and  ?

   The main reason I am on this list is because electronics hobbiests are an endangered spiecies.

Doug Hale

Mark Kinsler wrote:

Show quoted text

We generally do not normally work in units of electrical charge, so the concept seems a bit strange.  The easiest way to deal with this is to use the relationship I=Q/t, which says that current is the rate at which charge is delivered.  So we rewrite this definition as Q=It, which says that we can calculate the charge we've placed on a capacitor's plates by multiplying the current through the capacitor by the time we've applied that current.  Result: to determine charge, you need a watch and an ammeter.

If we charge a capacitor to a voltage V by pushing a charge Q (=It) onto the plates, we will find that the relationship between the voltage and the charge is Q/V=C, which is the definition of capacitance.  Note that we can thus measure the capacitance of a capacitor with an ammeter, a watch, and a voltmeter.

Now watch this:  We've charged the capacitor to voltage V.  We rewrite the definition of capacitance Q/V=C ---> V=Q/C.  This predicts what will happen if we reduce the capacitance by sliding the dielectric out from between the plates.  Yup: the voltage will go up, because we won't be reducing the charge on the plates.  C goes down, V goes up.  You can do the same thing by just pulling the plates apart.

The opposite will happen if we increase the capacitance of a capacitor.  If we were to charge an air dielectric capacitor to a voltage of 700 volts and then slip a piece of glass between the plates such that the glass completely fills the space between the plates, we'd find that the voltage has dropped to 100 volts.  Thus we say that glass has a relative dielectric constant of seven.  The dielectric constant of air and vacuum are equal for most practical purposes.

M Kinsler
512 E Mulberry St. Lancaster, Ohio USA 740 687 6368

_________________________________________________________________
Get your FREE download of MSN Explorer at 

To unsubscribe from this group, send an email to:
Electronics_101-unsubscribe@...

Your use of Yahoo! Groups is subject to

[email protected]

#1026

Now watch this:  We've charged the capacitor to voltage V.  We rewrite the 
definition of capacitance Q/V=C ---> V=Q/C.  This predicts what will happen 
if we reduce the capacitance by sliding the dielectric out from between the 
plates.  Yup: the voltage will go up, because we won't be reducing the 
charge on the plates.  C goes down, V goes up.  You can do the same thing by 
just pulling the plates apart.Voltage is just the potential between plates, kind of like weight.  I 
assume the work we do moving the dielectric goes into changing 
the potential.

The opposite will happen if we increase the capacitance of a capacitor.  If 
we were to charge an air dielectric capacitor to a voltage of 700 volts and 
then slip a piece of glass between the plates such that the glass completely 
fills the space between the plates, we'd find that the voltage has dropped 
to 100 volts.  Thus we say that glass has a relative dielectric constant of 
seven.  The dielectric constant of air and vacuum are equal for most 
practical purposes.As I recall, when a dielectric is placed between the plates, an 
opposing field is setup inside the dielectric.  I guess that opposing 
field is what is reducing the voltage on the plates.

Paul

Jim Purcell

#1031

Doug,

    The main reason I am on this list is because electronics hobbiests
are an endangered spiecies.Do you think so? I hope not. I used to think that radio Hams were too, but
I think that while their ranks may be reduce (maybe they aren't actually)
the ones who remain are just as dedicated. Maybe  Electronics Hobbyists
are like that too.

Jim

开云体育

capacitors and dielectrics