10.2 Electric Field Intensity

An electric field is a region of space around an electric charge or a group of charges within which an electric force is exerted on other charged objects. This field concept, introduced by Michael Faraday, provides a way to describe how electric forces are transmitted through space.

1. The Electric Field and Force

A source charge ($Q$) creates an electric field in the space surrounding it. When another charge, often called a "test charge" ($q$), is placed in this field, it experiences an electric force ($\vec{F}$). The electric field is the intermediary that communicates the force from the source charge to the test charge.

Scalar and Vector Quantities→

2. Electric Field Intensity ($\vec{E}$)

Electric field intensity (or simply the electric field) is a vector quantity that describes the strength and direction of the electric field at a specific point.

Definition: The electric field intensity at a point is defined as the electrostatic force ($\vec{F}$) experienced by a unit positive test charge ($q$) placed at that point.

Formula: $\vec{E}=\frac{\vec{F}}{q}$

• Magnitude: The magnitude of the electric field ($E$) represents the force per unit charge.
• Direction: The direction of $\vec{E}$ at a point is the same as the direction of the force that would be exerted on a positive test charge placed at that point.
  • Electric fields point away from positive source charges.
  • Electric fields point towards negative source charges.

3. Electric Field of a Point Charge

We can derive the formula for the electric field created by a single point source charge ($Q$) using Coulomb's Law.

1. The force ($F$) exerted by a source charge $Q$ on a test charge $q$ at a distance $r$ is: $\vec{F}=k\frac{Qq}{r^2}\hat{r}$
2. Using the definition of the electric field, $\vec{E}=\vec{F}/q$, we divide the force by the test charge $q$: $\vec{E}=\frac{1}{q}\left(k\frac{Qq}{r^2}\hat{r}\right)$
3. This gives the formula for the electric field intensity due to a point charge $Q$: $\vec{E}=k\frac{Q}{r^2}\hat{r}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}\hat{r}$ where $\hat{r}$ is a unit vector pointing from the source charge $Q$ towards the field point. The electric field strength decreases with the square of the distance from the source charge.

4. Units of Electric Field Intensity

From the defining equation ($E=F/q$), the SI unit of electric field intensity is Newtons per Coulomb (N C${}^{-1}$). It is also commonly expressed in Volts per metre (V m${}^{-1}$), since: $1{\text{V m}}^{-1}=1{\text{N C}}^{-1}$

Derived Units→

5. Example Calculation

Problem: A point charge of $q=9\mu \mathrm{C}$ is located at the centre of a sphere with radius $r=3.0\mathrm{m}$. Find the electric field intensity at the surface of the sphere.

Solution:

1. Formula: $E=k\frac{|q|}{r^2}$
2. Values:
   • $k=9\times 10^9\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-2}$
   • $q=9\mu \mathrm{C}=9\times 10^{-6}\mathrm{C}$
   • $r=3.0\mathrm{m}$
3. Calculation: $E=(9\times 10^9)\frac{9\times 10^{-6}}{(3.0{)}^2}=(9\times 10^9)\frac{9\times 10^{-6}}{9}=9\times 10^3\mathrm{N}{\mathrm{C}}^{-1}$

Result: The electric field intensity at the surface is $9\times 10^3\mathrm{N}{\mathrm{C}}^{-1}$, directed radially outward.