Electric field Intensity (Definition) and Electric field Intensity due to point charge

Definition of Electric Field Intensity:

The force acting on the per unit test charge in electric field is called the Electric field intensity. It is represented by $'E'$.

Let us consider that a test-charged particle of $q_{0}$ Coulomb is placed at a point in the electric field and a force $F$ acting on them so the electric field intensity at that point

$ \overrightarrow{E}=\frac{\overrightarrow{F}}{q_{0}}$

SI Unit:$\quad Newton/Coulomb$ $ (N/C)$ $\quad Kg-m^{2}/sec^{3} A$

Dimension: $\left [ML^{2}T^{-3}A^{-1} \right ]$

Physical Significance of Electric Field:

The force experienced by a charge is different at different points in space. So electric field intensity also varies from point to point. In general, Electric field intensity is not a single vector quantity but it is a set of infinite vector and each point in space have a unique electric field intensity. So electric field is an example of the vector field.

Electric Field Intensity due to a Point Charge:

Let us consider that an isolated point charge of $+q$ Coulomb is placed at a point $O$ in a medium whose dielectric constant is $K$. If a test charge particle of $q_{0}$ Coulomb is placed at a point $P$ in the electric field at a distance $r$ from point $O$. So the electric field intensity (magnitude form) at point $P$
Electric field intensity due to a point charge
Electric field intensity due to a point charge
$E=\frac{F}{q_{0}}\qquad (1)$

According to Coulomb's Law:

$F=\frac{1}{4\pi \varepsilon _{0}K}\frac{qq_{0}}{r^{2}}\qquad (2)$

From equation $(1)$ and equation $(2)$, we can write

$E=\frac{1}{4\pi \varepsilon _{0}K}\frac{q}{r^{2}} \qquad (3)$

For air or vacuum $K=1$, Then from equation $(3)$

$E=\frac{1}{4\pi \varepsilon _{0}}\frac{q}{r^{2}} \qquad (4)$

Where

$\frac{1}{4\pi \varepsilon _{0}}=9\times10^{9} N-m^{2}/C^{2}$

From equation $(4)$

$E=9\times 10^{9} \frac{q}{r^{2}}\qquad (5)$

In Vector Notation:

$\overrightarrow{E}=\frac{1}{4\pi \varepsilon _{0}K}\frac{q}{r^{2}}r\hat{}$

If a system containing $'n'$ point charge $q_{1},q_{2},q_{3},..........q_{n}$ then electric field intensity due to the system of charge particle will be equal to the vector sum of the intensities i.e.

$\overrightarrow{E}=\overrightarrow{E_{1}}+\overrightarrow{E_{2}}+\overrightarrow{E_{3}}+....+\overrightarrow{E_{n}}$

$\overrightarrow{E}=\frac{1}{4\pi \varepsilon _{0}K}\sum_{i=0}^{n}\frac{q_{i}}{r_{i}^{2}}r_{i}\hat{}$

Where $r_{i}$ is the distance from point $'P'$ to charge $q_{i}$.

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