 # Electric Field Due to Point Charge  View Notes

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.

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 experiences a force of attraction or repulsion.

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

So, 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:

$F = \frac{k|q_{1}|q_{2}|}{r^{2}}$

Where, $k = \frac{1}{4\pi \epsilon_{0}} = 9\times 10^{9} N m^{2} C^{-1}$

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 is 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:

 $\overrightarrow{E}(\overrightarrow{r}) = \overrightarrow{F} (\overrightarrow{r})$/q₀

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 N/C.

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.

From Fig.1,

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,

$F = \frac{1}{4\pi \epsilon_{0}} Q \frac{q_{0}}{r^{2}} (r)$

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

We know,

$\overrightarrow{E}(\overrightarrow{r}) = \overrightarrow{F} (\overrightarrow{r})$/q₀

Therefore,

 $\overrightarrow{E} = \frac{1}{4\pi \epsilon_{0}} \frac{Q}{r^{2}} (r)$

### Derivation of Electric Field Due to a Point Charge

Suppose the point charge +Q is located at A, where OA = $r_{1}$.

To calculate the electric field intensity (E) at B, where OB = $r_{2}$.

Let’s look at Fig.2,

From Fig.3:

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

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

= 1/4πε₀ q₁q₂ (r₁₂➙)/r₁₂³

 $\overrightarrow{F} = \frac{1}{\frac{4 \pi \epsilon_{0} q_{1}q_{2}}{|\overrightarrow{r_{2}} - \overrightarrow{r_{1}}|^{3} . (\overrightarrow{r_{2}} - \overrightarrow{r_{1}})}}$

Where, $AB = \overrightarrow{r_{12}} = \overrightarrow{r_{2}} - \overrightarrow{r_{1}}$

As,

$\overrightarrow{E} = \frac{\overrightarrow{F}}{q_{2}}$

Therefore,

 $\overrightarrow{F} = \frac{1}{\frac{4 \pi \epsilon_{0} q_{1}}{|\overrightarrow{r_{2}} - \overrightarrow{r_{1}}|^{3} . (\overrightarrow{r_{2}} - \overrightarrow{r_{1}})}}$

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.

Let’s say charge $Q_{1}, Q_{2}$ . . . Qₙ are placed in vacuum at positions $\overrightarrow{r_{1}}, \overrightarrow{r_{2}}, . . . . . \overrightarrow{r_{n}}$ respectively.

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

$\overrightarrow{F} = \frac{1}{4\pi \epsilon_{0}q_{0}} \sum_{i=1}^{i=n} \frac{Q_{i}}{|\overrightarrow{r} - \overrightarrow{r_{i}}|^{3} . |\overrightarrow{r} - \overrightarrow{r_{i}}|}$

As, $\overrightarrow{E} = \frac{\overrightarrow{F}}{q_{0}}$

Therefore,

 $\overrightarrow{F} = \frac{1}{4\pi \epsilon_{0}q_{0}} \sum_{i=1}^{i=n} \frac{Q_{i}}{|\overrightarrow{r} - \overrightarrow{r_{i}}|^{3} . |\overrightarrow{r} - \overrightarrow{r_{i}}|}$

Putting  1/4πε₀ = k

 $\overrightarrow{E} = kQ_{1}/r_{1}^{2} + kQ_{2}/r_{2}^{2} + . . . + kQ_{n}/r_{n}^{2}$

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

Q1: How is the Electric Field Generated?

Ans: The electric field is mathematically defined as a vector field that associates to each point in space the electrostatic force per unit of charge exerted on a unit positive test charge at rest at that point.

They are generated by electric charges, and charge configurations such as capacitors or by varying magnetic fields.

Q2: Why do Electric Field Lines Never Cross Each Other?

Ans: Electric field lines always point in one direction.

When two lines intersect each other, the tangents drawn at that point indicate two directions of electric field lines. This means the electric field has two directions at the same point. This is impossible. Therefore, electric field lines never cross each other.

Q3: Can Electric Fields Exist in a Vacuum?

Ans: No, electric fields and electric charges cannot exist in a vacuum because all the charge-carriers are massive particles or holes in a crystal of massive particles (atoms).

Q4: Can a Spark Occur Between Two Charges in a Vacuum?

Ans:  No spark will occur between two charges in a vacuum as the vacuum is not a good conductor of electricity. Even if we pass electricity through it as a stream of charged particles, there won’t be a spark.

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