[Physics Class Notes] on Electric Field Due to Point Charge Pdf for Exam

When a glass rod is rubbed with a piece of silk, the rod acquires the property of attracting objects like bits of paper, etc towards it.

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This happens due to the discharge of electric charges by rubbing of insulating surfaces.

Electric charge is a property that accompanies fundamental particles, wherever they exist.

When an electric charge q₀ is held in the vicinity of another charge Q, q₀ either experience a force of attraction or repulsion.

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We say that this force is set up due to the electric field around the charge Q.

Therefore, we can say that the electric field of charge Q as space by virtue of which the presence of charge Q modifies the space around itself leading to the generation of force F on any charge q₀ held in this space, given by:

                                          [F = frac{k|Q|q_|}{r^{2}}]

Here, from the above figure, we have the following parameters

r = The separation between source charge and test charge

Q = source charge,

q₁ = test charge, and

[k = frac{1}{4pi epsilon_{0}} = 9times 10^{9} N m^{2} C^{-1}]

        

The electric field due to a given electric charge Q is defined as the space around the charge in which electrostatic force of attraction or repulsion due to the charge Q can be experienced by another charge q.                      

Electric Field Due to a Point Charge Formula

The concept of the field was firstly introduced by Faraday.

The electric field intensity at any point is the strength of the electric field at that point.

It is defined as the force experienced by a unit positive charge placed at a particular point.

Here, if force acting on this unit positive charge +q₀ at a point r, then electric field intensity is given by:

[overrightarrow{E}({r}) = frac {overrightarrow{F}{(r)}}{q_o}]

Hence, E is a vector quantity and is in the direction of the force and along the direction in which the test charge +q tends to move.

Its unit is [frac {N} {C}].

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The electric field for +q₀ is directed radially outwards from the charge while for – q₀, it will be radially directed inwards.

Electric Field Due to a Point Charge Example

Suppose we have to calculate the electric field intensity or strength at any point P due to a point charge Q at O.

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From Fig.1 we have the following,

Here, OP = r.

Now, consider a small positive charge q₀ at P.

According to Coulomb’s law,  the force of interaction between the charges q₀ and Q at P is,                   

                        

                     [F = frac{1}{4pi epsilon_{0}} frac{Qq_{0}}{r^{2}}]

             

Where r is a unit vector directed from Q towards q₀.

We know, 

[overrightarrow{E}({r}) = frac { overrightarrow{F}(r)} {q_o}]

Therefore,                 

                                  

[overrightarrow{E} = frac{1}{4pi epsilon_{0}} / r^2 (r)]

Derivation of Electric Field Due to a Point Charge

Suppose the point charge +Q is located at A, where OA = r1.

To calculate the electric field intensity (E) at B, where OB = r2.

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From fig.2, we have:

According to Coulomb’s law, the force on a small test charge q2 at B is,

                                   

[F = frac{1}{4 pi epsilon_{0}} frac{q_{1}q_{2}(r_{12})}{r_{12^2}}]

                                            

[frac{1}{4 pi epsilon_{0}} frac{q_{1}q_{2}(r_{12})}{r_{12^3}}]

                    

[overrightarrow{F} = frac{1} {4pi epsilon_{0}}{frac{ q_{1}q_{2}}{|overrightarrow{r_{2}} – overrightarrow{r_{1}}|^{3} . (overrightarrow{r_{2}} – overrightarrow{r_{1}})}}]

Here, [AB = overrightarrow{r_{12}} = overrightarrow{r_{2}} – overrightarrow{r_{1}}]

As,            [overrightarrow{E} =  frac{overrightarrow{F}}{q_{2}}]

Therefore, 

[overrightarrow{F} = frac{1} {4 pi epsilon_{0}}{frac{ q_{1}}{|overrightarrow{r_{2}} – overrightarrow{r_{1}}|^{3} . (overrightarrow{r_{2}} – overrightarrow{r_{1}})}}]  

Hence,  E is produced along AB.  

Electric Field Due to a System of Point Charges    

The electric field intensity at any point due to a system or group of charges is equal to the vector sum of electric field intensities due to individual charges at the same point.

Now, we would do the vector sum of electric field intensities:

[overrightarrow{E}  = overrightarrow{E_{1}} + overrightarrow{E_{2}} + overrightarrow{E_{3}} + … + overrightarrow{E_{n}}]

      

[overrightarrow{E} = frac{1}{4 pi epsilon_{0}} sum_{i=1}^{i=n} frac{widehat{Q_{i}}}{r_{i}^{2}} . (r_{i})]

Here, 

[r_{i}] is the distance of the point P from the ith charge [Q_{i}] and [r_{i}] is a unit vector directed from [widehat{Q_{i}}] to the point P.

r_i is a unit vector directed from Q_i to the point P.                                                          

Let’s say charge Q₁, Q₂…Q_n are placed in vacuum at positions r₁, r₂,….,rₙ respectively.                 

                                                         

The net forces at P are the vector sum of forces due to individual charges, given by,

[overrightarrow{F} = frac{1}{4pi epsilon_{0}} q_{0} sum_{i=1}^{i=n} frac{overrightarrow Q_{i}}{|overrightarrow{r} – overrightarrow{r_{i}}|^{3} . |overrightarrow{r} – overrightarrow{r_{i}}|}]]                                   

  As, [overrightarrow{E} = frac{overrightarrow{F}}{q_{0}}]

 

 Therefore, 

[overrightarrow{E} = frac{1}{4pi epsilon_{0}} sum_{i=1}^{i=n} frac{overrightarrow Q_{i}}{|overrightarrow{r} – overrightarrow{r_{i}}|^{3} . |overrightarrow{r} – overrightarrow{r_{i}}|}]]

Putting  [frac {1}{4 pi epsilon_{0}}]     
= k         

         

[overrightarrow{E} = k frac {Q_{1}} {r_{1^2}} + k frac {Q_{2}}{r_{2^2}} + . . . + k frac {Q_{n}} {r_{n^2}}]  

Hence, we obtained a formula for the electric field due to a system of point charges.