By Richard Carter

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**Example text**

Thus, for the geometry shown above it is sufficient to find the solution for one quadrant of the problem. 30) is entered into all the other cells. When symmetry has been used to reduce the size of the problem the formulae in the cells along symmetry boundaries make use of the fact that the potentials on either side of the boundary are equal. d) The formulae in the cells are then applied repeatedly ( a process known as iteration ) until the numbers in the cells cease to change. To do this the calculation options of the spreadsheet must be set to permit iteration.

Such a device is a source of electric power because it does work in driving the electric current around the circuit against the circuit resistance. As the charges move around the circuit from one terminal of the source to the other, they lose potential energy as it is converted into heat. 16)) This equation remains true near a current-carrying circuit provided that the path of integration does not pass through the generator. But if the line integral of the electric field is evaluated around the circuit from one terminal of the source to the other, the result is not zero.

The type of charge carrier; 2. the material in which it moves; 3. the temperature of the material. It can also depend on 4. the strength of the electric field; 5. the orientation of the crystal axes to the field. For many materials ȝ is a scalar quantity which varies only with temperature and this will be assumed here. 2) where n is the density of charge carriers and q the charge on each. com 49 Electromagnetism for Electronic Engineers 3. Steady electric currents where ı is the conductivity of the material, measured in siemens per metre.