# Electric field energy: experiments and formulas

When considering the energy of an electric field, one should study its accumulation and consumption. Energy accumulators are electrocondensers. With small dimensions, such a device is able to concentrate a large amount of energy.

When studying capacitors, it becomes easier to understand the electrostatic laws and capabilities of modern devices. These are, for example, the well-known digital multimers, with which they carry out measurements in picofarads. First, the parameters should be estimated using electrostatic methods, and after that - using a multimer.

## Electric capacity of the extended conductor

The study of this device gives a better understanding of the question of what is the energy of an electric field. Conductors are able to accumulate and save charges. This property is called electrical capacitance.

To understand the dependence of the potential of an elongated conductor on the charge, it is necessary to measure the potentials of a charged body. It is convenient to do this on the ground.

An electrometer with a hollow conducting ball and a grounded body is used as an electrostatic voltmeter and the potential of the body relative to the ground is measured.

The probe ball touches an electrical source, thus transferring a charge into it. In this case, the voltmeter will show the presence of a certain potential.

Repeating the experience, it can be concluded that the charge-to-potential ratio is constant.

By changing the hollow ball to another one and having done the same experiments, if the voltmeter shows large values in comparison with the previous ones, it can be concluded that the second ball has a smaller capacity.

In the international SI system, the unit of measurement of electrical capacitance is farad.

## Experience with a spherical conductor

If in a medium with a dielectric constant we take a spherical conductor, where the potential at infinity is zero, then the potential in the sphere with charge will be equal to Q / 4ПƐ˳ƐR, and the capacitance C = 4ПƐ˳ƐR,

It turns out that the electric capacity of the elongated ball is proportional to its radius.

From the experiments, it follows that the bodies are considered elongated if the surrounding bodies do not cause a significant redistribution of the charge in them.

## Capacitor

A capacitor is made of two identical parallel plates and an electrometer is connected to it, which will work as a voltmeter. A conducting sphere is brought to its rod. The plate is charged by transferring the charge from the ebonite stick. Then the voltmeter will indicate the presence of voltage that has arisen between the plates.

Transferring equal charges inside the hollow sphere, increase the instrument readings. Therefore, the capacity of the plates will be as follows: C = q / U, capable of operating as a capacitor accumulating a charge of electricity (where q is the charge of one of the plates).

## Capacity of a flat capacitor

The capacitance of a plane capacitor is C = ε̥ε / d, where d is the distance between the plates.

The formula can be confirmed by experiment. A flat capacitor is assembled, the plates are charged and connected to a voltmeter. Without changing the charge, they change other indicators, watching the device at this time. The readings will be inversely proportional to the capacitance: U = q / C – 1 / C.

Making the distance between the plates more, we will observe an increase in voltage. By shifting the plates in parallel and increasing the area, we obtain a reduction in voltage, and the capacity will increase.If a dielectric is placed in the gap between the plates, the readings of the voltmeter will decrease.

Since in the course of the experiment the value of the charge was not changed, it turns out that the capacitor capacitance is directly proportional to the overlap of the plates and inversely proportional to d.

## Parallel and serial capacitor connection

When the devices are connected in parallel, the capacitances of the devices and their voltages have the same values, and the charges are different. The total charge is equal to their sum separately.

With a serial connection, connect a voltmeter having a hollow sphere. For one plate of the first capacitor, a positive charge is given, then the other plate will become negative, and when connected to the conductor of the second device - positive. Then both capacitors will receive identical charges, and their voltages will have different values.

As a result, the capacity here will be determined by the formula: 1 / C = 1 / C1 + 1 / C2

## Energy of flat and arbitrary capacitor

A charge is applied to the plate, having a value at which the potential difference between the plates becomes equal to U. Then the intensity will be equal to E = U / d, where d is the distance between objects.

One of the plates is in the electric field of the other, where the intensity is E / 2. Then the force of attraction to the other plate will be f = qE / 2. The potential energy of the electric field of a charge is equal to the operation of this field when the plates approach each other.

Substituting a series of values, we obtain that the field energy is W = qU / 2 = q² / 2C = CU² / 2.

This formula is suitable for any capacitor. The total work of the field is A = 1 / 2qU.

The same thing happens if you apply an elongated conductor instead of a capacitor.

## Determination of energy experimentally

Measurement of the energy of the device produced by thermal action. A metal helix is placed in the tube, closed with a stopper with a tube in which there is a drop of water. Get a gas thermometer. A capacitor is connected to the coil, and an electrometer with a ball hollow inside is connected in parallel.

The capacitor is charged with balls, and then discharged through a spiral. You will notice the movement of the drop in the tube.

After cooling the air and moving the drop to the initial position, the voltage increases. The drop will move several values higher. The capacitor is changed to a large capacity twice. By charging it to the initial level, you can observe the movement doubled.

## Electric field energy density

They give energy such that the capacitor does not have values, and only the values characterizing the field are taken into account. In this case, the energy of the electric field per unit volume must be calculated.

As a result of substitutions, the energy density is obtained: ω = W / V = ε̥εΕ² / 2, that is, it is proportional to the square of the intensity.

## Charge interaction energy or energy in an electric field

So, to charge a capacitor, work is needed to overcome the forces of electrostatic attraction between different charges when they are separated. Due to this, there will be a reserve of potential energy.

To charge any body, work is also necessary, in this case, to overcome electrostatic repulsion between like charges.

Taking a solitary conductor, we charge q. The potential of the field at infinity will be zero, and the potential of the conductor - φ (q). To transfer a small ∆q charge, work is required:

∆A = φ (q) ∆q.

Work on charging a solitary conductor is determined by the formula:

A = W = 1/2 φ (Q) Q = 1 / 2C (φ (Q)) ²

The question of where energy is stored is answered in two ways. According to one of them, this is the energy of interaction of charges on the conductor, and according to another, the energy of the electric field is obtained, since it is distributed in the surrounding space.

Which answer from these two to give preference is a personal decision of each student. However, it should be noted that when studying variable fields, only the second option becomes possible, where the energy is associated with an electric field.