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Continuous charge distributions spread over a line, or a surface, or volume, are introduced. The electric field due to a continuous charge distribution is obtained by dividing the charge distributions into infinitesimal elements, and summing over electric fields due to the small elements. The computation can often be simplified by making use of symmetries of the problem.
1. Volume charge distribution
Frequently it is useful to think that charges are continuously distributed over
a volume (or on a surface or on a curve).So we introduce a function $\rho$
called charge density. Consider a small volume \(dV\) located at point \(P\).
Total charge inside $dV = \rho(P)dV$
Where $\rho(P)$ is the value of $\rho$ computed at \(P\) and $\rho$ is the volume
charge density or the charge density per unit volume.
To compute the electric field due to a volume charge distribution :
1. Divide the charge distribution into small volumes.
2. Find the electric field due to each volume.
3. Use the superposition principle to get he total $\bar{E}$ as sum (or integral
over \(V\)) over all the volume elements.
2. Line charge distribution / surface charge distribution
One can similarly have a continuous charge distribution over a curve. So we
introduce a function $\lambda$ called linear charge density which is equal to
charge per unit length.
Total charge on a small piece of length $dl = \lambda dl $
Similarly, surface charge density is defined by $\sigma$ which is equal to
charge per unit area.
Total charge on a small area $dS = \sigma dS$
Electric field due to charges along a curve (or on a surface) is found by :
1. Dividing the curve (or surface) into small pieces.
2. Finding $\bar{E}$ due to one piece.
3. Use superposition principle, total electric field is then obtained by summing
over all pieces (or integrating over the curve or surfaces)
3. Examples
1. Let three equal charges be placed on the corners od an equilateral triangle. Prove that the electric field at the center of the triangle is zero.
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4727:Diamond Point
