In this article, we shall discuss the electric field due to charged particles at a point and the field direction, and several facts.

**The electric field at a point is the resultant field generated by all the charged particles surrounding that point and the intensity of the field is directly proportional to the source charge and the distance of separation of the point from the source.**

**How to Find Electric Field at a Point?**

The electric field produced by the charged particle can either be attractive or repulsive depending upon the charge of the particle.

**The electric field at any point around this region formed by the charged particle is directly proportional to the charge that it carries and inversely proportional to the distance of separation between the charge and the point in consideration.**

The electric field at a point due to the presence of a charge q_{1} is simply given by the relation

Where q_{1} is a charge producing the electric field

r is a distance separating the charge and the point

Incase if there is a charge present at a point P then we know that the electric force between the two charged particles is

Where q_{1} is a charged particle

And q_{2} is a particle at a point P in an electric field formed by particle q_{1}

r is a distance separating two particles

The same is depicted in the below diagram

The direction of the electric field is shown in the diagram, since the particle at point P is oppositely charged the electric force is an attractive force.

Then the electric field formed by the particle q_{1} at a point P is

This is a formula to calculate the electric field at any point present in the field developed by the charged particle.

**Problem 1: What is the electric field at a point due to the charge of 5C which is 5cm away?**

**Given:**

q_{1}=5C

r=5cm=0.05m

The electric field due to charge q_{1}=5C is

9*10^{9}*5C/(0.05)^{2}

45*10^{9}/0.0025

18*10^{12}N/C

The electric field at a point is 18*10^{12}N/C

**How to Find Net Electric Field at a Point?**

The net electric field at a point is a sum of all the electric fields exerting at a point.

**The net electric field can be calculated by adding all the electric fields acting at a point, the electric fields can be attractive or repulsive based on the charge that generates the electric field.**

Consider the following diagram showing differently charged particles q_{1}, q_{2}, q_{3}, and q_{4} surrounded by the point P separated at different distances r_{1}, r_{2}, r_{3}, and r_{4} respectively from the point.

Now here, the electric field due to charge q_{1} is

The same way, the electric field due to charge q_{2} is

The electric field due to charge q_{3} is

The electric field due to charge q_{4} is

Then the net electric field at point P is

If there are ‘n’ numbers of charges, then the net electric field at a point due to all the charges is

**How to Find Electric Field Strength at a Point?**

The electric field strength is a field intensity and potential of a field at a point.

**The electrostatic force can be calculated as the ratio of the electrostatic force and the charge on which it the exerting the force or else the charge produces the electric field at a certain point separated by some distance.**

**Problem 2: What is the electric field strength at a point separated at a distance of 0.25 m from the charge of +2C?**

**Given:** q=+2C

r= 0.25 m

We have,

=9*10^{9}*2/(0.25)^{2}

9*10^{9}*2/0.0625

228*10^{9}N/C

Hence the electric field at a point 0.25m far away from the charge of +2C is 228*10^{9}N/C

**How to Find Electric Field Intensity at a Point?**

**It can be calculated as the ratio of the electric force experienced at a point per unit charge of the particle and is given by the relation **E=F/q

The more the electrostatic force imposed on the charges or at a point by the source particle, the more will be the intensity of the electric field space generated by the charged particle. The intensity of the field will be a maximum when the spacing between the point and the source will be a minimum and if the source charge carries the higher charge.

**How to Find Direction of Electric Field at a Point?**

We can find the direction of the electric field at a point by introducing the test charge in the electric field.

**Basically, the direction of the positively charged particle is radially outward whereas, that of the negatively charged particle the direction of the field is radially inward.**

On introducing the point charge in the electric field region, the charge will show sudden drift and align itself in the direction of the field this indicates the direction of the electric field produced by the source charge.

If we place the positive test charge in the field, then the direction of the electric field is as shown in the below diagram:-

And that of the negative point charge, the direction of the electric field is radiating inwards as shown below:-

If we place two oppositely charge carriers in an electric space then the direction of the field will be running from the positively charged particle to the negative charge carrier.

If there is two charges having similar charges are placed in a field, then the repulsive force will act on each of the charges. Suppose we have two positive charges, then the repulsive force will exert push force on each other.

**How to Find Magnitude of Electric Field at a Point?**

The magnitude of the electric field at a point is the net electric force experienced on the unit charge at that point.

**The magnitude of an electric field is calculated using a formula **

**and the magnitude of the field is always positive irrespective of the sign of the charge.**

**What are the magnitude and the direction of the electric field at a point away from the source charge at a distance of 15cm having a charge of -15mC?**

**Given:** q=-15mC

r=15cm=0.15m

=9*10^{9}Nm^{2}C^{-2*}[-15*10^{-6}]/(0.15)^{2}

=135*10^{3}/0.0225

=6*10^{6}N/C

The magnitude of the electric field is 6*10^{6}N/C

**Electric Field at a Point on Equatorial Line**

The equatorial line is a line perpendicular to the axial line of the dipole connecting the two oppositely charged carriers.

Consider a point ‘P’ on the equatorial line, the electric field at point P due to charge –q is

And the electric field at point P due to charge +q is

The magnitude of both the electric field is equal,

We can calculate the net electric field at a point P by applying the Parallelogram Law of vector addition.

[E_{1}]=[E_{2}]

E=2E_{1}Cosθ—-(5)

Substituting value for ‘E’ we have,

From triangle APO, we find the value of Cosθ as

Cosθ=l/√r^{2}=l^{2}

Using this in the above equation,

p=2ql

For r>>>l,

**Electric Field at a Point on Axial Line of Dipole**

The dipole is formed due to the separation of the oppositely charges at some distance.

**The axial point is the center point between the two charges forming electric dipoles, our aim is the find the electric field on this axial line joining the point at the middle of the two charges.**

Consider two charges +q and –q and an axial point between the two located at point ‘O’. The distance between the two charges be ‘2l’. Let ‘p’ be the point on the axial line.

The electric field intensity at point P due to charge +q is

E=1/4π∈_{0}*q/(r-l)^{2}

And the electric field intensity at point P due to charge -q is

Hence, the net electric field at a point P on the axial line of dipole E=E_{1}+E_{2}

q/4π∈_{0}*q(1/(r-l)^{2}-1/(r+l)^{2})

q/4π∈_{0}(4rl/(r^{2}-l^{2})^{2})

We know that

Electric moment

Therefore,

If r>>>l, then

**Electric Field at a Point on Equatorial Plane**

Consider an equatorial plane standing at an axial point ‘O’. The magnitude of the electric field at a point ‘P’ on the plane is equal due to the charges +q and –q.

The net electric field at a point is

From eqn (6), we know that

The total electric field is opposite to the electric dipole and hence the net electric field is negative.

For large distance i.e. r>>a,

Hence, the electric field at equatorial plane is

**Electric Field at a Point on the Axis of Charged Ring**

Consider a uniformly charged ring of radius ‘r’ and a small charged element dq on the ring. Let P be the point lying on the center axis of the charged ring at a distance ‘l’ from its center. Let θ be the angle formed on the axis and a line joining point P and the charge element.

**The net electric field is due to all the charges around the ring. The field perpendicular to the axis is zero hence the only component of the electric field that comes into consideration is the x-component.**

The electric field element

From the diagram

Hence,

Integrating this equation

This is equal to the electric field at a point on the axis running from the center of the charged ring.

For large distances r>>>l,

This equation gives the electric field at a point on the axis of the charged ring that has a large radius.

**Electric Field at a Point due to a Point Charge**

Consider a source charge Q producing the electric field

Let q be the test charge placed in this field at a distance ‘r’ from the source charge.

The electric force between the two charges now produced is

The electric field due to a point charge is E=F/q

Which is equal to

Q can be positive or negative depending upon the charge that it carries.

**Electric Field at a Point due to Two Charges**

**If there are two charges Q _{1} and Q_{2} separated by some distance ‘r’ then the electric force between the two is **

The electric field due to charge Q_{1} at point P is

The electric field due to charge Q_{2} at point P is

The net electric field at a point is

The electric field at a point depends upon the number of charges surrounding it and the electric force exerting on that point.

**Electric Field Intensity at a Point in Between Two Parallel Sheets**

Consider two parallel sheets having charge densities +σ and –σ separated by some distance.

**The electric field lines will be running from the positively charged plate to the negatively charged plate. The electric field is perpendicular to the plane sheet and the magnitude of the electric field is **

Let P be the point between the two parallel sheets. The magnitude of the electric field is equal, and in the same direction as shown in the figure between the two plates hence the net electric field at point P is

**This is the electric field intensity at a point between the two charged plates. At any point outside this charge parallel sheet, the electric field intensity is zero.**

**Frequently Asked Questions**

**If the net electrostatic force between the two charges +3C and -2C are placed at points A and B respectively separated at a distance of 10 fm then what is the distance of a point from A where the electric field strength is zero?**

**Given:** q_{1}=+3C

q_{2}=-2C

d=10 fm

d-s=(10-s) fm

Let the electric field produced by charge q_{1},E_{b} and the electric field produced by charge q_{2 }be E_{b}

The point at which the electric field strength is zero is

Solving this equation using quadratic formula

Separation can’t be negative, hence eliminating another part and considering only the positive term of the equation, we find

Hence, the distance of a point from A where the electric field strength is zero is

d-s= 10-4.5=5.5 fm

Read more about Are Electric Field Lines Perpendicular?

**Also Read:**

- Does oil conduct electricity
- Does gold conduct electricity
- Electric field example
- Can fusion generate electricity
- Does lead conduct electricity
- Does carbon conduct electricity
- Does sodium conduct electricity
- Does granite conduct electricity
- Does bismuth conduct electricity
- Is static electricity neutral

Hi, I’m Akshita Mapari. I have done M.Sc. in Physics. I have worked on projects like Numerical modeling of winds and waves during cyclone, Physics of toys and mechanized thrill machines in amusement park based on Classical Mechanics. I have pursued a course on Arduino and have accomplished some mini projects on Arduino UNO. I always like to explore new zones in the field of science. I personally believe that learning is more enthusiastic when learnt with creativity. Apart from this, I like to read, travel, strumming on guitar, identifying rocks and strata, photography and playing chess.