Investigating the capacitance of a parallel report

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Investigating the capacitance of a parallel-plate capacitor using a reed switch


To investigate the factors which affect the capacitance of a parallel-plate capacitor using a reed switch


Reed switch        

Signal generator        

Capacitor plates        

Polythene spacers

Polythene sheet        

Battery box with 4 cells        


Resistance substitution box        

Light-beam galvanometer

Standard mass

Connecting leads                


Through the series of experiments, we would like to verify the following theories

  1. Charge and Applied Potential Difference

With the increase of voltage supplied by battery, the current is increased. Moreover, in a V-I graph, the slope m is the proportional constant fC. As we can obtain the generate our desired frequency f by a signal generator, we can obtain the capacitance.            .

  1. Effect of Plate Separation and Area of Overlap

If the capacitor is charged up by voltage V, the charge density of the capacitor plates is given by

The electric field inside the plates is given by

With the increase of plate separation, the capacitance is decreased. As Q= CV, with constant voltage supply, the number of charges stored is decreased.

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With the decrease of area of overlap, the capacitance is also decreased. Hence, the number of charges stored is also decreased.

  1. Relative permittivity

It is known that by putting a dielectric between the plates could increase the capacitance. A dielectric is a material that can be polarized by an electric field. When a dielectric material is placed in a uniform electric field, one surface will contain many positive ends of molecules and the other surface will contain many negative ends. When a dielectric is placed between charged plates, the polarization of the medium produces an electric field opposing ...

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