Talha's Physics Academy
Electric Field and Electric Field Intensity
Video Lecture
Watch the complete video lecture below to understand electric fields, electric field intensity, and the derivation for an isolated point charge.
Electric Field and Electric Field Intensity
Electric Field
Electric Field Intensity
If a test charge experiences a force $\vec{F}$, then the intensity of the electric field $\vec{E}$ is given by:
$$\vec{E} = \frac{\vec{F}}{q_0}$$
Where $q_0$ is the positive test charge ($q_0 \to 0$).
- Vector Nature: Electric field intensity is a vector quantity.
- Direction: Its direction is away from the source charge if the charge is positive, and towards the source charge if the charge is negative.
- SI Unit: The SI unit of electric field intensity is Newton per Coulomb ($\text{N/C}$).
Electric Field Intensity Near an Isolated Point Charge
Consider a very small positive test charge $q_0$ placed at a distance $r$ from a point charge $q$. The magnitude of the electrostatic force $\vec{F}$ on $q_0$ due to the charge $q$ is given by Coulomb's Law:
$$F = \frac{1}{4\pi \varepsilon_0} \frac{q q_0}{r^2}$$
The electric field intensity $E$ at the position of $q_0$ due to $q$ is given by substituting the force equation into the field definition:
$$E = \frac{F}{q_0}$$
$$E = \frac{1}{q_0} \left( \frac{1}{4\pi \varepsilon_0} \frac{q q_0}{r^2} \right)$$
$$E = \frac{1}{4\pi \varepsilon_0} \frac{q}{r^2}$$
Thus, the electric field intensity due to a point charge varies inversely with the square of the distance ($r^2$) from the charge and is independent of the test charge $q_0$.
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