Title : C2 Investigating the capacitance of a capacitor
I. Objective :
* To investigate the factor which affect the capacitance of a parallel-plate capacitor using a reed switch.
* Reed Switch
* Signal generator
* Capacitor Plates (1 pair)
* Polythene spacers
* Low voltage power supply
* Variable resistance
* Light-beam galvanometer
* Standard mass (eg. 100g )
* Connecting leads
* Drawing board
* Vernier calipers for measuring thickness of spacer
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A capacitor is a device which can store charge. It consists of two sheets of conductor separated by a layer of insulator called dielectric. The capacitance of a capacitor is defined as the charge stored per unit potential difference applied to the capacitor :
In the experiment, a reed switch allows the capacitor to be charged up and discharged rapidly :
The reed switch consists of three small metal strips (F,N and K) inside a glass capsule which is surrounded by a coil. F and K are made of steel but N is non-magnetizable material. In normal condition, K (called the reed) is in contact with N. When a current passes through the coil, F and K become oppositely magnetized so that K is attracted towards F and makes contact with it. When the current falls, K springs back into contact with N.
If the frequency at which the reed switch is operated is f, the charging and discharging process will be repeated f times per second, and the charge Q on the capacitor is delivered to the microammeter at the same rate.
The capacitor is fully charged up and discharged every time, the total quantity of charge passes through the microammeter in a second is :
This gives the size of theoretical current I.
The capacitance of the capacitor can be estimated from
Hence, the reading of the microammeter reflects the capacitance of the parallel-plate capacitor.
For a pair of plates are oppositely charged and arranged so that they are parallel to
one another and only a small distance apart, then the electric field between them will be uniform except at their edges. If the edge effect is ignored the value of the
electric field strength between the plates is
Since the field is uniform, we also have
Equating these two values of E gives
This gives the capacitance of the parallel plate capacitor, C, to be
Since capacitance is defined as C =Q/V , the charge on a capacitor is directly proportional to the potential difference at which the capacitor is charged up. The graph of charge Q against potential difference V gives the slope representing the capacitance of a parallel-plate capacitor.
1. The galvanometer was set to the shorted position and the light sport was moved to off-set zero. The sensitivity of was set at X1.0 and it was recorded 2. The circuit was connected as shown. A drawing board was placed on the bench and the parallel plate capacitor horizontally was put on top. The capacitor plates with 4 polythene spacers placed at the corners 3. The frequency and voltage of the signal generator was adjusted such that a sound was heard and the spot on the light-beam galvanometer was deflected 4. Asdasd
5. A variable resistor was connected in series with the galvanometer to protect it from damage, which may result from an accidental connection of the d.c. supply to the galvanometer. 6. A CRO was connected across the variable resistance to monitor the discharging current pulse. The variable resistor was adjusted to a maximum value such that zero was reached by each pulse before the next one. A) Charge and applied p.d.
1. The signal generator was set to the...
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