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How to find the magnitude of an electric field?

A field is a means of thinking about and visualizing the force that surrounds any charged object and acts on another charged object at a distance, even if there is no obvious physical contact between these two objects.

Electric field E due to set of charges at any point is the force experienced by a unit positive test charge placed at that point.

The units of electric field are N/C or V/m.

Electric field E is a vector quantity meaning it has both

  • magnitude and
  • direction

In this article we will learn how to find the magnitude of an electric field.

Magnitude of electric field created by a charge

Let us now first look at finding magnitude of electric field created by a charge.

To find electric field due to a single charge we make use of Coulomb’s Law.

If a point charge q is at a distance r from the charge q then it will experience a force

F=14πϵ0qqˆrr2

Electric field at this point is given by relation

E=Fq=14πϵ0qˆrr2

This is electric field at a distance r from a point charge q and ˆr is the unit vector along the direction of electric field.

Above relation defining electric field at a distance r tels about both magnitude and direction of the field.

Magnitude of electric field would be

|E|=14πϵ0|q|r2

In above equation you could notice the missing ˆr part. This ˆr tells us about the direction of the electric field. Since electric force is a central force and we have defined electric field using Coulombs law we can conclude that electric field acts along the line joining the charge q (source point) and field point at which it is being measured.

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This vector This ˆr is known as unit vector of our displacement vector r and is given by relation

ˆr=rr

Where r=|r| which is the distance between our source point and the field point.

Magnitude of electric field Formula

Thus, the magnitude of electric field due to a point charge is given by relation

E=Fq=14πϵ0qr2

It is important to note here that the magnitude of E depends on the charge q wgich produces the electric field not on the value of test charge q.

Solved Example 1

Find the magnitude of electric field at a distance 0.2m from a charge of 4nC.

Solution

Step 1 – Figure out what is requited

The question asked us to find magnitude of electric field

Step – 2 Figure out information given in the question.

We are provided the magnitude of the charge as well as the distance between the field point and the charge.

Step 3- Find out the way to solve problem

We will use the relation

E=kQr2

To find the magnitude of electric field.

Step 4- Solve the problem

E=kQr2

where

k=14πϵ0=9.0×109Nm/C2

So,

E=(9.0×109)4×109.22=900N/C

Magnitude of electric field due to multiple charges

Magnitude of electric field due multiple charges can be calculated using superposition principle.

It states that “The total electric field at a point P is the vector sum of the fields at P due to each point charge in the charge distribution.”

Magnitude of electric field

To explain this further let us consider the figure which shows two electric charges q1 and q2 and we have to find net electric field at point P due to these two charges.

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According to principle of superposition

E=Fq=E1+E2

If we have knowledge about the magnitude of charges and distance of point P from both these charges then we can use relation

E=kQr2

Where k=14πϵ0=9.0×109Nm/C2

To estimate the resultant field, we must calculate the electric field for each charge separately and then add them together.

Let us now understand this with the help of a problem.

Solved Example 2

Two point charges qA=3μC and qB=3μC are placed at 20 cm apart in vacuum. What would be the electric field at the mid point O of the line AB joining the two charges?

Solution

Step- 1 Figure out what is requited

The question asked us to find magnitude of electric field at the center of line joining both the charges.

Step- 2 Figure out information given in the question.

We are provided the magnitude of the charge as well as the distance between both the charges

qA=3μC=3×106C

qB=3μC=3×106C

r=20cm=0.2m we have to convert r given in cm to m.

Let 0 be the mid point of the line as shown below in the figure

how to find Magnitude of electric field

Then,

OA=OB=r2=0.22=0.1m

Step- 3 Find out the way to solve problem

We will use the relation

E=kQr2

To find the magnitude of electric field.

Step 4- Solve the problem

First we will calculate electric field at point O due to qA

EA=k|qA|(OA)2=9×109×3×106(0.1)2=2.7×106N/C

Acting along OB

Next we will calculate electric field at point O due to qB

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EB=k|qB|(OB)2=9×109×3×106(0.1)2=2.7×106N/C

Acting along OB

Now since EA and EB are acting along same direction the angle between them is θ=0radians and coso=1

|E|=E2A+E2B+2E1E2cosθ

=E2A+E2B+2E1E2

=|EA|+|EB|

Therefore net magnitude of electric field at point O due to two charges qA and qB would be,

E=EA+EB=2.7×106+2.7×106=5.4×106N/C

Note:- If EA and EB acts along opposite direction the angle between them is θ=πradians and in that case |E|=|EA||EB|