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)
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)
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. 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 \([MLT^{-3}A^{-1}]\).
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
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.
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, \[\vec{F}=\frac{q_0}{4\pi \varepsilon _0}\sum_{i=1}^n{\frac{q_i\left( \vec{r}-\vec{r}_i \right)}{\left| \vec{r}-\vec{r}_i \right|^3}}\\Therefore,\\\vec{E}=\frac{\vec{F}}{q_0}=\frac{1}{4\pi \varepsilon _0}\sum_{i=1}^n{\frac{q_i\left( \vec{r}-\vec{r}_i \right)}{\left| \vec{r}-\vec{r}_i \right|^3}}\]
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