Investigating the Capacitance of a Parallel-Plate.

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Investigating the Capacitance of a Parallel-Plate

Capacitor Using a Reed Switch

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

Apparatus: One reed switch, one signal generator, one pair of capacitor plates, polythene spacers, polythene sheet, one battery box, one voltmeter, one resistance substitution box, one light beam galvanometer, one 100g standard mass, connecting leads

THEORIES

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

. Charge and Applied Potential Difference

As

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 .

Capacitance is the amount of charges which can be stored per unit voltage applied to the capacitor. The unit is defined as farad (F).

2.

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 where A is the area of the capacitor plate

The electric field inside the plates is given by

where is the permittivity in vacuum

Since

And,

and

As ,

and

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

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

3.

Effect of Dielectric

It is known that by putting a dielectric between the plates could increase the capacitance.

A dielectric is a material that can be polarised 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 polarisation of the medium produces an electric field opposing the field of the charges on the plate.

As and ,

As the capacitance is inversely proportional to the electric field between the plates, and the presence of the dielectric reduces the effective electric field, the capacitance of the parallel plate is increased.

The permittivity is a characteristic of space, and the relative permittivity is a way to characterise the reduction in effective field because of the polarisation of the dielectric.
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Assume the permittivity of the dielectric is.

The relative permittivity is defined as , where is the permittivity of free space.

Therefore, the capacitance of a capacitor with a dielectric is given by

As ,

As ,

Also, for constant electric field strength, area of overlap and frequency

So,

As ,

EXPERIMENTS

We have set up a circuit using the apparatus as shown below.

The battery box contains 4 batteries, which allows us to change the electromotive force (e.m.f.) in our investigations on the relationships ...

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