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.
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)
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 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.
We use term ‘electric field intensity’ or ‘electric field strength’ to describe the strength of electric field due to a charge.
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.
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
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 \(P\).
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.
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
\[\left| \vec{E} \right|=\frac{1}{4\pi \varepsilon _0}\frac{q}{r_{12}^{2}}=9\times 10^9\frac{q}{r_{12}^{2}}\]
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, \[\vec{F}=\frac{q_0}{4\pi \varepsilon _0}\sum_{i=1}^n{\frac{q_i\left( \vec{r}-\vec{r}_i \right)}{\left| \vec{r}-\vec{r}_i \right|^3}}\\Therefore,\\\vec{E}=\frac{\vec{F}}{q_0}=\frac{1}{4\pi \varepsilon _0}\sum_{i=1}^n{\frac{q_i\left( \vec{r}-\vec{r}_i \right)}{\left| \vec{r}-\vec{r}_i \right|^3}}\]
link to this page by copying the following textThanks for visiting our website.
DISCLOSURE: THIS PAGE MAY CONTAIN AFFILIATE LINKS, MEANING I GET A COMMISSION IF YOU DECIDE TO MAKE A PURCHASE THROUGH MY LINKS, AT NO COST TO YOU. PLEASE READ MY DISCLOSURE FOR MORE INFO.