In the previous discussions of electrostatics several definitions were discussed. These were Coulomb's Law, Electric Field Intensity (E), Electric Flux Density (D) and Potential/ Voltage (V).
Here Capacitance will be defined after briefly adding a bit more knowledge of potential/voltage.
Previously Potential/Voltage was discussed (See: Electrostatics -Potential: The Definition Of Voltage)
By looking at the formula for the voltage (see previous thread) it can be seen that voltage between two points is equal to the Voltage difference between two points. Or rather a voltage exists between two points, as in the example between point a and point b. To put this in a different context, there is a 'Potential Difference'.
If the voltage was to be calculated between a point a and zero (0) then we have what is called absolute potential. A physical example of these definitions would be: Potential Difference:- Take a voltage source of say 4 volts and a second voltage source of 2 volts, connect a voltmeter across the two output terminals from each source and the reading will be 2 volts - as this is a potential defference which in theis case is 4 - 2 = 2 volts. Also remember that voltage/potential exists between a place of higher energy (here 4) to a place of lower energy (here 2). Generally speaking take a voltage a and a voltage b then the voltage out would be a - b (or the highest value minus the lowest value - potential difference).
Absolute Potential:- Next take a voltage source and a ground/ 0 volt output, when measuring the voltage this time, say the voltage source is 6 volts, then the output voltage would be 6 - 0 = 6 volts.
Generalls speaking take a voltage source a and with reference to 0 volts we get: a - 0 = a, Absolute Potential.
The Relationship Between Electric Field Strength (E) and Voltage (V).
You may have notived that if we take the voltage formula (absolute in this case): V = Q/4πεod, and divide this formula by the distance d (used d in this example to represent distance, could use any letter), we have the Electric Field Intensity (E): E = Q/4πεod2. (if the squared value doesn't show here, it might look like d2 by default, take it to be d squared)
Lets do the calculation to show this: V = (Q/4πεod) * 1/d = Q/4πεod * d. So therefore we can say that th electric field intensity (E) is equal to V/d - In a statement then, the Electic Field Intensity (E) is equal to volts per meter.
Similarly by manipulating this formula (transposing for V) we see that the voltage (V) is equal to E * d (the product of Electric Field Strength and distance d).
Capacitance
Capacitance is well known to be the ratio( quotient/division) of Charge (Q) and Voltage (V).
In a formula then: C = Q/V.
But we know from the discussion above that V is equal to E * d, lets put this information into the formula C = Q/V and see what happens.
C = Q / E * dl . Now expanding E from what we already know we get. C = Q / (Q/4πεod2). d
which is equal to: C =Q/(Qd/4πεod2) now we can cancel out (but please write these formulas down in order to make sure that you can do these simplifications for yourself as it looks kind of messy on a computer screen). Well first of all the d on the numerator of the fraction (top bit in Qd/4πεod2 ) will cancel out with one of the d's on the numerator (bottom bit) giving thus far; C = Q/(Q/4πεod), now we have a quotient (division) and as when dividing fractions we will flip the part Q/4πεod upside down to give us a product hence; C = Q * (4πεod/Q) which gives Q4πεod/Q, now the Q's cancel out finally giving C = 4πεod.
So from this definition of Capacitance we can see that Capacitance is defined not by Charge (Q) or voltage (V), but simply by the geometry (Shape) of an object, and the materials used (εo).
NOTE: The εo represents the permittivity of free space,air,Vacuum, but when we are using other materials such as glass, mica e.t.c. we use εr along with εo as we compare how good the material εr is to εo which gives ε = εo * εr.
So we could write C = 4πεd.
Capacitance in Parallel Plate Capacitors will be defined in another thread - please digest this information in order to comprehend Electrostatics and its uses.
Thanks
Nirvana.
Here Capacitance will be defined after briefly adding a bit more knowledge of potential/voltage.
Previously Potential/Voltage was discussed (See: Electrostatics -Potential: The Definition Of Voltage)
By looking at the formula for the voltage (see previous thread) it can be seen that voltage between two points is equal to the Voltage difference between two points. Or rather a voltage exists between two points, as in the example between point a and point b. To put this in a different context, there is a 'Potential Difference'.
If the voltage was to be calculated between a point a and zero (0) then we have what is called absolute potential. A physical example of these definitions would be: Potential Difference:- Take a voltage source of say 4 volts and a second voltage source of 2 volts, connect a voltmeter across the two output terminals from each source and the reading will be 2 volts - as this is a potential defference which in theis case is 4 - 2 = 2 volts. Also remember that voltage/potential exists between a place of higher energy (here 4) to a place of lower energy (here 2). Generally speaking take a voltage a and a voltage b then the voltage out would be a - b (or the highest value minus the lowest value - potential difference).
Absolute Potential:- Next take a voltage source and a ground/ 0 volt output, when measuring the voltage this time, say the voltage source is 6 volts, then the output voltage would be 6 - 0 = 6 volts.
Generalls speaking take a voltage source a and with reference to 0 volts we get: a - 0 = a, Absolute Potential.
The Relationship Between Electric Field Strength (E) and Voltage (V).
You may have notived that if we take the voltage formula (absolute in this case): V = Q/4πεod, and divide this formula by the distance d (used d in this example to represent distance, could use any letter), we have the Electric Field Intensity (E): E = Q/4πεod2. (if the squared value doesn't show here, it might look like d2 by default, take it to be d squared)
Lets do the calculation to show this: V = (Q/4πεod) * 1/d = Q/4πεod * d. So therefore we can say that th electric field intensity (E) is equal to V/d - In a statement then, the Electic Field Intensity (E) is equal to volts per meter.
Similarly by manipulating this formula (transposing for V) we see that the voltage (V) is equal to E * d (the product of Electric Field Strength and distance d).
Capacitance
Capacitance is well known to be the ratio( quotient/division) of Charge (Q) and Voltage (V).
In a formula then: C = Q/V.
But we know from the discussion above that V is equal to E * d, lets put this information into the formula C = Q/V and see what happens.
C = Q / E * dl . Now expanding E from what we already know we get. C = Q / (Q/4πεod2). d
which is equal to: C =Q/(Qd/4πεod2) now we can cancel out (but please write these formulas down in order to make sure that you can do these simplifications for yourself as it looks kind of messy on a computer screen). Well first of all the d on the numerator of the fraction (top bit in Qd/4πεod2 ) will cancel out with one of the d's on the numerator (bottom bit) giving thus far; C = Q/(Q/4πεod), now we have a quotient (division) and as when dividing fractions we will flip the part Q/4πεod upside down to give us a product hence; C = Q * (4πεod/Q) which gives Q4πεod/Q, now the Q's cancel out finally giving C = 4πεod.
So from this definition of Capacitance we can see that Capacitance is defined not by Charge (Q) or voltage (V), but simply by the geometry (Shape) of an object, and the materials used (εo).
NOTE: The εo represents the permittivity of free space,air,Vacuum, but when we are using other materials such as glass, mica e.t.c. we use εr along with εo as we compare how good the material εr is to εo which gives ε = εo * εr.
So we could write C = 4πεd.
Capacitance in Parallel Plate Capacitors will be defined in another thread - please digest this information in order to comprehend Electrostatics and its uses.
Thanks
Nirvana.
