Looking back over the Electrostatics described earlier, we have enough tools to analyze the behaviour of a Parallel Plate Capacitor, and in particular it's Capacitance (C).
Consider the two plates in a Parallel Plate Capacitor , we will assume that these plates are an exact copy of each other. Lets say that the top plate is connected to the positive terminal of a battery whilst the bottom plate is connected to the negative terminal of a battery.
To calculate the total charge stored on the top and bottom plate we would take the amount of charge and divide it by the area of the plate - this would tell us how much charge is spread over the plate.
The top plate in this case would have Q/A amount of charge on its surface. The bottom plate would have
- Q/A charge on its surface, the negative sign shows negative charge from negative terminal of the battery.
Keeping the above in mind I'd like to point out a useful concept that is the Flux density (D) = ε E , Therefore
E = D/ε, Now what might not have been discussed earlier it that the electric flux density (D) is how dense the charge is or how much charge there is in a given area so D = Q/A.
With that in mind E becomes; E = Q/εA.
Ok now back to Capacitance, well as stated in another thread Capacitance is the ratio of Charge (Q) to Voltage (V) or
C = Q/V. Now also from previous discussions is was found that V = E. d, and E (From above) = Q/εA therefore putting it all together we get: C = Q/V = Q / (Qd/εA) flippling side Qd/εA to give a product as you would do in fractions to give:
C = Q * (εA/Qd) = QεA/Qd, now the Q's cancel out giving the result, C = εA/d.
Here we see that the Capacitance then is not dependant by the amount of charge there is or voltage, but by the Geometry (Shape) of the component itself.
Nirvana.
Consider the two plates in a Parallel Plate Capacitor , we will assume that these plates are an exact copy of each other. Lets say that the top plate is connected to the positive terminal of a battery whilst the bottom plate is connected to the negative terminal of a battery.
To calculate the total charge stored on the top and bottom plate we would take the amount of charge and divide it by the area of the plate - this would tell us how much charge is spread over the plate.
The top plate in this case would have Q/A amount of charge on its surface. The bottom plate would have
- Q/A charge on its surface, the negative sign shows negative charge from negative terminal of the battery.
Keeping the above in mind I'd like to point out a useful concept that is the Flux density (D) = ε E , Therefore
E = D/ε, Now what might not have been discussed earlier it that the electric flux density (D) is how dense the charge is or how much charge there is in a given area so D = Q/A.
With that in mind E becomes; E = Q/εA.
Ok now back to Capacitance, well as stated in another thread Capacitance is the ratio of Charge (Q) to Voltage (V) or
C = Q/V. Now also from previous discussions is was found that V = E. d, and E (From above) = Q/εA therefore putting it all together we get: C = Q/V = Q / (Qd/εA) flippling side Qd/εA to give a product as you would do in fractions to give:
C = Q * (εA/Qd) = QεA/Qd, now the Q's cancel out giving the result, C = εA/d.
Here we see that the Capacitance then is not dependant by the amount of charge there is or voltage, but by the Geometry (Shape) of the component itself.
Nirvana.