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See http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html, section on field from disk.mysci said:
Looks right.mysci said:
In fact, I don't understand although I can deduce it.haruspex said:
dz indicates an arbitrarily thin slice. Such a thin slice of a sphere is effectively a very short cylinder. This is the basis of calculus.mysci said:
An E-field, or electric field, is a measure of the force that a charged particle would experience at a given point in space. In the case of a continuous charge distribution, the E-field is a vector quantity that describes the electric force at every point in the distribution.
One of the main problems is determining the exact value of the E-field at any given point, as it is dependent on the charge distribution and can be difficult to calculate. Another problem is understanding the behavior of the E-field in complex charge distributions, as it can be affected by the geometry and arrangement of charges.
The E-field is inversely proportional to the square of the distance from a continuous charge distribution. This means that as the distance increases, the strength of the E-field decreases. However, the exact relationship between distance and E-field strength depends on the charge distribution and its geometry.
In some cases, yes. For simple charge distributions, such as a point charge or a uniformly charged sphere, the E-field can be calculated using Coulomb's law or Gauss's law. However, for more complex charge distributions, numerical methods may be necessary to obtain an accurate value for the E-field.
The E-field can exert a force on charged particles, causing them to accelerate and change their trajectory. This is the basis for many applications in electrical engineering and particle physics, such as particle accelerators. The behavior of the E-field in a continuous charge distribution can also affect the stability of charged particles in an electric field.
