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Saturday, December 8, 2012

Electrostatics

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If a conductor has a deficit of electrons, it will exhibit a net positive charge. If, on the other hand, it has a surplus of electrons, it will exhibit a net negative charge. An imbalance in charge can be produced by friction (removing or depositing electrons using materials such as silk and fur, respectively) or induction (by attracting or repelling electrons using a second body which is, respectively, positively or negatively charged).

Force between charges

Coulomb’s Law states that, if charged bodies exist at two points, the force of attraction (if the charges are of opposite polarity) or repulsion (if the charges have the same polarity) will be proportional to the product of the magnitude of the charges divided by the square of their distance apart. Thus:
where Q1 and Q2 are the charges present at the two points (in Coulombs), r the distance separating the two points (in metres), F is the force (in Newtons), and k is a constant depending upon the medium in which the charges exist. In vacuum or ‘free space’, 
 




Combining the two previous equations gives:
Electric fields

The force exerted on a charged particle is a manifestation of the existence of an electric field. The electric field defines the direction and magnitude of a force on a charged object. The field  itself is invisible to the human eye but can be drawn by constructing lines which indicate the motion of a free positive charge within the field; the number of field lines in a particular region being used to indicate the relative strength of the field at the point in question.





Figures 1.7 and 1.8 show the electric fields between charges of the same and opposite polarity while Fig. 1.9 shows the field which exists between two charged parallel plates.








Electric field strength

The strength of an electric field (5) is proportional to the applied potential difference and inversely  proportional to the distance between the two conductors. The electric field strength is given by:   

E = V / d

where E is the electric field strength (V/m), V is the applied potential difference (V) and d is the distance (m).

Example

Two parallel conductors are separated by a distance of 25 mm. Determine the electric field strength if they are fed from a 600 V d.c. supply.

Solution

The electric field strength will be given by:
 


Permittivity

The amount of charge produced on the two plates shown in Fig. 1.9 for a given applied voltage will depend not only on the physical dimensions but also on the insulating dielectric material that appears between the plates. Such materials need to have a very high value of resistivity (they must not conduct charge) coupled with an ability to withstand high voltages without breaking down.

A more practical arrangement is shown in Fig. 1.10. In this arrangement the ratio of charge, Q, to potential difference, V, is given by the relationship:
where A is the surface area of the plates (in m), d is the separation  (in m), and  E is a constant for the dielectric material known as the  absolute permittivity of the material (sometimes also referred to as the dielectric constant).

The absolute permittivity of a dielectric material is the product of the permittivity of free space ( E0) and the  relative permittivity ( Er) of the material. Thus:
The dielectric strength  of an insulating dielectric is the maximum electric field strength that can safely be applied to it before breakdown (conduction) occurs. Table 1  shows values of relative permittivity and dielectric strength for some common dielectric materials.

















Electromagnetism

When a current flows through a conductor a magnetic field is produced in the vicinity of the conductor. The magnetic field is invisible but its presence can be detected using a compass needle (which will deflect from its normal North South position). If two current-carrying conductors are placed in the vicinity of one another, the fields will interact with one another and the conductors will experience a force of attraction or repulsion (depending upon the relative direction of the two currents). 

Force between two current-carrying conductors

The mutual force which exists between two parallel current-carrying conductors will be proportional to the product of the currents in the two conductors and the length of the conductors but inversely proportional to their separation. Thus:

where  I1 and  I2 are the currents in the two conductors (in Amps), l is the parallel length of the conductors (in metres), d is the distance separating the two conductors (in metres),  F is the force (in Newtons), and  k is a constant depending upon the medium in which the charges exist.  In vacuum or ‘free space’,



Combining the two previous equations gives:

Magnetic fields

The field surrounding a straight current-carrying conductor is shown in Fig. 1.11. The magnetic field defines the direction of motion of a free North pole within the field. In the case of Fig. 1.11, the lines of flux are concentric and the direction of the field determined by the direction of current flow) is given by the right-hand rule.


Magnetic field strength

The strength of a magnetic field is a measure of the density of the flux at any particular point. In the case of Fig. 1.11, the field strength will be proportional to the applied current and inversely proportional to the perpendicular distance from the conductor. Thus:




where B is the magnetic flux density (in Tesla), I is the current (in amperes), d is the distance from the conductor (in metres), and k is a constant.

Assuming that the medium is vacuum or ‘free space’, the density of the magnetic flux will be given by:



The flux density is also equal to the total flux divided by the area of the field. Thus:




In order to increase the strength of the field, a conductor may be shaped into a loop (Fig. 1.12) or coiled to form a solenoid (Fig. 1.13). Note, in the latter case, how the field pattern is exactly the same as that which surrounds a bar magnet. We will see


















 Example

Determine the flux density produced at a distance of 50 mm from a straight wire carrying a current of 20 A.












Example

A flux density of 2.5 mT is developed in free space over an area of 20 cm2. Determine the total flux.
Solution






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