Capacitance is defined as the ratio of the Charge to the voltage. . For this lab parallel plate devices are considered and the capacitance of such devices is calculated by the formula
Where k is the dielectric constant for the material between the plates, for a it’s 1, and is the Permittivity of the vacuum which is a constant, 8.854*10^-12 F/m. The constants are accepted measure of the ability to accept an electric field of the material between the plates. It’s important to consider the material and the geometry of the device. In the formula A is the area of the plates and d is the distance between them. Traditionally your told d must be very small compared to the area and many calculations are done assuming the plates are infinite, but a lab is real life and we’re going to settle for a set of circular plates 18cm in diameter separated by distances varying from 1mm to 50mm. Another problem is our inability to measure the capacitance, the electrometer only measures voltage. We can substitute into and we have
Where, since for any given charge, everything is constant and V is proportional to d. There are some complicating factors when dealing with capacitors. While still considered passive, they have a transient response. When power is first applied to them they act as a wire and current flows, but of course a capacitor isn’t a wire and as charge flows into them they become charged at which point they behave as an open circuit. If the circuit only has a capacitor, this is almost instantaneous and the capacitor goes from a short circuit to an open circuit. If there is resistance in the circuit, this happens in a short time determined by the values of the resistance and capacitance of the system. The other complicating factor is how to measure it. To determine the capacitance you have to charge the capacitor, but the measuring device will affect the measurement by draining off charge. There are meters for such measurement and just as with resistors a network of capacitors can be combined into a single capacitor.
In a series circuit, the sum of the voltages must add up to the total voltage, by KVL and conservation of energy. The charge on each capacitor has to equal the total charge and you can use to determine the formula for the total capacitance in a series circuit.
In a parallel circuit, the voltage across each capacitor has to be the same as the total voltage supplied and the sum of the individual charges must equal the total charge, and now you can use to find that the total capacitance is the sum of the individual capacitances.
|Distance between plates (mm)||Voltage in V||Vplates=(1+644d)Vsystem|
The graph given for the v plate is linear and the intercept is zero.
The calculated charge is 5.853*10^-12 coulombs.
In the video the change in dielectric caused the voltage to drop. Is that correct? Yes it is Explain why? This is because of decrease in electrostatic field between the plates that causes the voltage to drop because of charge cancellation by the dielectric.
Precise measurements are difficult in these cases, but from the voltages recorded and the formulas in the introduction, calculate the dielectric constant for the paper used in the experiment. The plates are 18cm in diameter and the separation is 8cm.
Voltages recorded are 27.5 and 23.6
For alternating current;
The capacitors values in series
C1 = 2.0 uf, C2= 3.1 uf, C3=4.4 uf and C4 = 6.8 uf.
PLEASE SEE ATTACHED PDF FILE FOR CALCULATIONS
Capacitors in parallel
PLEASE SEE ATTACHED PDF FOR CALCULATIONS
Combination of both series and parallel
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SEE ATTACHED PDF FOR CALCULATION
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