 # Electric field

From the study of electric forces and Coulomb’s Law we came to know that there exists a force between two charged particles when they are kept at some distance apart from each other. Now a question arises how we can explain the concept of this action-at-a-distance force. This is explained using a concept of electric field.

## What is Electric Field?

Electrical interaction between charged particles can be reformulated using the concept of electric field. To understand the concept consider the mutual repulsion of two positive charged bodies $A$ and $B$. Charged body $A$ has charge $q$ and charged body B has charge $q_0$ as shown in figure (a)

Now if we remove the charged body $B$ and label its position as point $P$ as shown in figure (b), then charged body $A$ is said to produce an electric field at that point (and at all other points in its vicinity)
When a body $B$ is placed at point $P$ and experiences force $F$, we explain it by a point of view that force is exerted on B by the field created due to charged body A not by body $A$ itself.
So, we can say that body $A$ sets up an electric field and the force on body $B$ is exerted by the field due to $A$.
An electric field is said to exists at a point if a force of electric origin is exerted on a stationary charged (test charge) placed at that point.

### Electric Field Definition

Electric field due to a charge is the space around the charge in which any other charge experiences a force of attraction or repulsion.
In figure (a) the charge $+q$ is called the source charge because it is producing the electric field.
The charge $+q_0$ is called the test charge. Test charge should be as small as possible so that its presence does not have any effect on the electric field due to source charge.

### Electric Field Intensity

We use term ‘electric field intensity’ or ‘electric field strength’ to describe the strength of electric field due to a charge.

#### Electric field intensity definition

The Electric field intensity at a point is defined as the force experienced by a unit positive charge placed at that point.
Now consider the figure given below where a charge $q$ is located at point $O$ in space. If $\vec{F}$ is the force acting on test charge $q_0$ placed at a point $P$ in an electric field then electric field at that point is $\vec{E} = \frac{\vec{F}}{q_0}$ or $\vec{F} = q_0\vec{E}$ Electric field is a vector quantity and since $\vec{F} = q_0\vec{E}$ the direction of $\vec{E}$ is same as the direction of $\vec{F}$. Unit of electric field is $(N.C^{-1})$ and the dimensions of electric field is $[MLT^{-3}A^{-1}]$.
It must be noted here that test charge we are considering here should be as small as possible we can make it so small so that it approaches zero or become infinitesimally small. In this case we can define electric field due to such infinitesimally small charge as $\vec{E}=\underset{q_0\rightarrow 0}{lim}\frac{\vec{F}}{q_0}$

### Physical Significance of electric field

• It is very important concept in understanding various electrostatic phenomenon.
• The space around every electric charge or electrically charged body is filled with an electric field thereby altering the space around it. This is the reason why electrostatic force like gravitational force is an action-at-a-distance force.
• Electric field should not be thought of as a kind of matter filled in space surrounding electric charge. It is a kind of aura or the distinctive atmosphere or quality that seems to surround and be generated by an electric charge.

## Calculation of Electric Field

In previous section we studied a method of measuring electric field in which we place a small test charge at the point, measure a force on it and take the ratio of force to the test charge.
Electric field at any point can be calculated using Coulomb's law if both magnitude and positions of all charges contributing to the field are known.
To find the magnitude of electric field at a point $P$, at a distance$r$ from the point charge q as shown below in the figure

Let us imagine a test charge $q_0$ to be placed at P. Now we find force on charge $q_0$ due to q through Coulomb's law. $\vec{F}=\frac{1}{4\pi \epsilon _0}\frac{qq_0}{r^2}\hat{r}$ Electric field at point P is $\vec{E}=\frac{\vec{F}}{q_0}=\frac{1}{q_0}\left( \frac{1}{4\pi \epsilon _0}\frac{qq_0}{r^2} \right) \hat{r}$ Or, $\vec{E}=\frac{1}{4\pi \epsilon _0}\frac{q}{r^2}\hat{r}$ And $\left| \vec{E} \right|=\frac{1}{4\pi \epsilon _0}\frac{q}{r^2}=9\times 10^9\frac{q}{r^2}$ Above expression gives the magnitude of electric field intensity due to charge $q$ at any point at a distance $r$ from it.
The direction of the field is away from the charge q if it is positive. Electric field for either a positive or negative charge in terms of unit vector $\hat{r}$ directed along line from charge q to point P is $\vec{E}=\frac{1}{4\pi \epsilon _0}\frac{q}{r^2}\hat{r}$ Here, $r$ is the distance from charge $q$ to point $P$.

## Electric Field in terms of position vectors

Now we will learn how to express electric field intensity due to a point charge in terms of position vectors.
For this consider a point charge $+q$ located at point $A$ as shown below in the figure.

From this figure we can see that $\overrightarrow{OA}=\vec{r}_1$ is the position vector of our source charge kept at point $A$. Our problem here is to find the electric field intensity at $P$ whose position vector is $\overrightarrow{OB}=\vec{r}_2$ due to our source charge $+q$ at point $A$.
Now from triangle law of vector addition $\overrightarrow{OA}+\overrightarrow{AP}=\overrightarrow{OP}\\\overrightarrow{AP}=\vec{r}_{12}=\overrightarrow{OP}-\overrightarrow{OA}\\\vec{r}_{12}=\vec{r}_2-\vec{r}_1$ If we place a infinitesimally small test charge $+q_0$ at point $P$ then force $\vec{F}$ acting on this charge $+q_0$ can be found using Coulomb’s Law and is given by, $\vec{F}=\frac{1}{4\pi \varepsilon _0}\frac{qq_0}{r_{12}^{2}}\hat{r}_{12}=\frac{1}{4\pi \varepsilon _0}\frac{qq_0}{r_{12}^{3}}\vec{r}_{12}\\\vec{F}=\frac{qq_0}{4\pi \varepsilon _0}\frac{\left( \vec{r}_2-\vec{r}_1 \right)}{\left| \vec{r}_2-\vec{r}_1 \right|^3}$ Therefore, $\vec{E}=\frac{\vec{F}}{\boldsymbol{q}_0}=\frac{q}{4\pi \varepsilon _0}\frac{\left( \vec{r}_2-\vec{r}_1 \right)}{\left| \vec{r}_2-\vec{r}_1 \right|^3}$ Electric field is directed along from A to P and its magnitude is $\left| \vec{E} \right|=\frac{1}{4\pi \varepsilon _0}\frac{q}{r_{12}^{2}}=9\times 10^9\frac{q}{r_{12}^{2}}$

## $\vec{E}$ due to a group of point charges

The resultant or net electric field intensity at a point due to a group of point charges can be calculated by applying the principle of superposition. To calculate $\vec{E}$ due to a group of point charges consider the number of point charges which are at positions $\vec{r}_1, \vec{r}_2, \vec{r}_3,.........., \vec{r}_n$ from point $P$. Our problem is to find electric field intensity at point $P$ whose position vector is $\vec{r}$. Now, from principle of superposition we have. $\vec{E}=\vec{E}_1+\vec{E}_2+\vec{E}_3+.........+\vec{E}_n\\where\\\vec{E}=resultant\,\,electric\,\,field\,\,at\,\,point\,\,P\\\vec{E}_1=Electric\,\,field\,\,intensity\,\,at\,\,P\,\,due\,\,to\,\,q_1\\\vec{E}_2=Electric\,\,field\,\,intensity\,\,at\,\,P\,\,due\,\,to\,\,q_2\,\,and\,\,so\,\,on.$ Again let us now imagine that a very small positive test charge $+q_0$ is placed at point $P$ then force on this test charge $+q_0$ is equal to vector sum of forces due to charges $q_1,\,\,q_2,\,\,q_3,\,\,..........q_n$ on $+q_0$. Therefore, $\vec{F}=\frac{q_0}{4\pi \varepsilon _0}\sum_{i=1}^n{\frac{q_i\left( \vec{r}-\vec{r}_i \right)}{\left| \vec{r}-\vec{r}_i \right|^3}}\\Therefore,\\\vec{E}=\frac{\vec{F}}{q_0}=\frac{1}{4\pi \varepsilon _0}\sum_{i=1}^n{\frac{q_i\left( \vec{r}-\vec{r}_i \right)}{\left| \vec{r}-\vec{r}_i \right|^3}}$ 