What is electric field?
An electric charge produces an electric field, which is a region of space around an electrically charged particle or object in which an electric charge would feel force. The electric field exists at all points in space and can be observed by bringing another charge into the electric field. However, the electric field can be approximated as zero for practical purposes if the charges are far enough from each other.
Electric fields are a vector quantity and can be visualized as arrows going toward or away from charges as shown in Fig.no.1. The lines are defined as pointing radially outward, away from a positive charge, or radially inward, toward a negative charge.
Fig.No.1: Left: Opposite Charges attract each other, Right: Same charges repel each other.
This electric field is represented by electric
flux lines, which are drawn to indicate the strength of the electric field
at any point around the charged body; that is, the denser the lines of
flux, the stronger the electric field. In Fig. 2, the electric field
strength is stronger at position a than at position b because the flux lines
are denser at a than at b. The symbol for electric flux is the Greek letter $\psi$ (psi). The flux per unit area (flux density) is represented by the capital letter D and is determined by
$$\bbox[10px,border:1px solid grey]{D = {\psi \over A}} \quad \text{(flux / unit area)} \tag{1}$$
Fig. 2: Flux distribution from an isolated positive charge.
The larger the charge Q in coulombs, the greater the number of flux
lines extending or terminating per unit area, independent of the surrounding medium. Twice the charge will produce twice the flux per unit
area. The two can therefore be equated:
$$ \bbox[10px,border:1px solid grey]{\psi = Q} \quad \text{(coulombs, C)} \tag{2}$$
By definition,
The electric field strength at a point is the force acting on a unit positive charge at that point; that is,
$$ \bbox[10px,border:1px solid grey]{E = {F \over Q}} \quad \text{(newtons/coulomb, N/C)} \tag{3}$$
The force exerted on a unit positive charge ($Q_2 = 1 C$), by a charge
$Q_1$, r meters away, as determined by Coulomb's law is
$$ F = {kQ_1 Q_2 \over r^2} = {k Q_1 (1) \over r^2} = {k Q_1 \over r^2} \tag{4}$$
Substituting this force F into Eq. (3) yields
$$ E = {F \over Q_2} = {k Q_1/r^2 \over 1} $$
$$ \bbox[10px,border:1px solid grey]{ E = {k Q/r^2} } \tag{5}$$
We can therefore conclude that the electric field strength at any point
distance r from a point charge of Q coulombs is directly proportional to
the magnitude of the charge and inversely proportional to the distance
squared from the charge. The squared term in the denominator will
result in a rapid decrease in the strength of the electric field with distance from the point charge.
Electric Field Related Questions
