Unit of Electric Field

Introduction

Charged particles (positive or negative) create an electric field around them. This electric field may be either attractive or repulsive based on the nature of the charged particles present. As we all know, the same charges (positive-positive and negative-negative) repel while the opposite charges (positive-negative) attract each other. The English alphabet “E” represents the magnitude/ intensity/ strength as well as the direction of this electric field. The unit of an electric field in the SI system is V/m or Volte per metre. The SI unit is equivalent to N/m. However, the unit of an electric field in the CGS system is statvolt cm^-1. Learn more about the electric field and its properties.

Coulomb’s Law

Coulomb’s law deals with the force of repulsion or attraction present between any two charged bodies. This law states that the force (attraction or repulsion) exerted by charged particles is inversely proportional to the square of the distance between them and is directly proportional to the product of these charges.

Thus, the derived formula from the above statement is as follows:

F = Kq₁q₂/r²

Here, K = Coulomb constant (≈ 8.988×10⁹ N⋅m²⋅C^-2)

F = Force (attraction or repulsion)

q₁ and q₂ = Magnitude of the charges

r = distance between the charges.

Define: Electric Field

Let’s understand the electric field by assuming that a point charge (Q) is present in a vacuum. It has an origin point termed O. Now, let’s assume that another point charge denoted as “q” is placed at point P. The distance between the points “O” and “P” is “r”. According to Coulomb’s law, q will receive a force by the charge Q. Thus, the charge Q will now be responsible for creating an electric field in its surroundings. Thus, the magnitude of this electric field is:

E = kQ/ r²

Here, K = Coulomb constant (≈ 8.988×10⁹ N⋅m²⋅C^-2)

E = Electric field

Q = Magnitude of the charge

This equation tells us about the value or magnitude of the present electric field.

Let’s use the vector “r” to represent the position of the charge “q”.  The representation of it in an equation is as follows:

F (r) = qE (r)

The equation shows that the experienced force ‘F’ will be the same as the product of the electric field (E) and the charge (q). Moreover, according to this equation, the unit of an electric field will be Newton/Coulomb or N/C.

Example: Suppose a 10 N force acts on a given charge of 4 μ C at point H. Evaluate the magnitude of the electric field at H.

Solution:

According to the question,

Force = 10 N and the charge (q)= 4 μ C

We can simply evaluate it by dividing the force by the given charge.

Thus, E = F(force) / q(charge)

= 10N / 4×10⁻⁶C

E = 2.5 × 10⁶ N/C.

Unit of Electric Field

The unit of an electric field in the SI system is V/m or Volte per metre. However, the unit of an electric field in the CGS system is statvolt cm⁻¹. Moreover, as stated above, the unit of an electric field can also be Newton/Coulomb or N/C.

Electric Field: System of Charges

Suppose there is an electric field with the various number of charges labelled as q₁, q₂, q₃, and q₄… All these charges have their position vectors as r₁, r₂, r_3, and r₄…, respectively. Their origin point is the same i.e. O.

The basic criteria in the system of charges will be the same as one charge. The force experienced by a single charge placed at a point without disorganizing other charges (q₁, q₂, q₃, q₄) will be the electric field. Let’s derive the equation with the help of Coulomb’s law along with the Superposition Principle, taking the vector “r” as the point “P”.

E₁ = k q₁/ r₁²

Similarly,

E₂ = k q₂/ r₂²

E₃ = k q₃/ r₃²

E₄ = k q₄/ r₄²

So the magnitude of the net electric field will be

E = E₁ + E₂ + E₃ + E₄

E = k(q₁/ r₁² + k q₂/ r₂² + q₃/ r₃² + k q₄/ r₄² )

The net electric field will be the sum of all electric fields.

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