Electric Field

The electric field stands as a fundamental concept in physics, defining the influence that electric charges exert on their surroundings. This field has both direction and magnitude . It guides the movement of charged entities, impacting everything from the spark of static electricity to the functionality of electronic devices Understanding electric fields will help you to understand how charge particles interact with each other and the surroundings. And guides various natural and technological phenomena.

What is an Electric Field?

Electric Field is a fundamental concept in physics, they are a physical field that surrounds an electrically charged particle. Charged particles exert attractive force when they have opposite charges and repulsive when they have the same charges. The electric field of a single charge describes its power to exert such forces on another charged object.

The electric field is a vector quantity which means it has both magnitude and direction.

Mathematical Expression for Electric Field

Mathematically electric field(E) is described at a given point is defined as the force (F) experienced by a test charge (q) placed at that point, divided by the magnitude of the test charge:

E=F/q

Direction Of Electric Field : The direction of electric field is taken as the direction of the force which is exerted on the positive charge. The electric field is radially outwards from the positive charge and radially inwards to the negative point charge, as shown in the picture given below:

Electric Field Lines

Key Terminologies Related to Electric Field

Now, let us see some of the important terminologies related to electric field that will be discussed further in the article:

• Electric Charge: An electric charge is a property of matter that causes two objects to repel or attract. It can be either negative or positive.
• Point charge: When discussing a body considerably smaller than the distance being considered, we disregard its size and refer to it as a point charge.
• Coulomb’s Law: Coulomb’s law states that the force between two point charges is directly proportional to the magnitudes of the charges and inversely proportional to the distance between the two charges.
• Mathematically : F = (k|q1q2|)/r2 ,where q1=first point charge ,q2 = second point charge.
• k = 8.988 * 10⁹ Nm2/C2 is Coulomb’s constant, r = the distance between two point charges.
• Gauss’s law : Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.( further discussed in detail in the article).
• Electric flux: Electric flux is defined as the total number of electric field lines passing through a specified area within a unit of time.

Properties of Electric Field Lines

Electric field lines have different properties. Some of the properties are provided below:

• Field lines never intersect each other.
• They are perpendicular to the surface charge.
• The strength of an electric field increases when the field lines are closer together, which indicates a stronger force. Conversely, when field lines move farther apart, the field weakens.
• The quantity of field lines directly proportional to the magnitude of the charge.
• These lines generally originate from positive charges and ends at negative charges, depicting the direction of the electric field.

Electric Field Calculation

Various methods are followed to calculate electric field are as followed:

To Calculate Electric Field Using Coulomb’s Law

According to Coulomb’s law, a force with electric charge q1 at position x1 exerts a force on a particle with charge q2 at position x0 of ,(equation-1)

F= 1/(4\piε0) .(q1.q2)/(x1-x0)2 r^1,0

where , r_1,0 is a unit vector in the direction from point x₁ to point x₀.

[Tex]\epsilon [/Tex]₀ is the electric constant, known as absolute permittivity of free space .

The electric field is force per unit charge,

formula for electric field is given in eqution-2.

E(x0) = F/q2 = 1/(4\piε0). q1 / (x1 – x0)2r^1,0

To Calculate Electric Field Using Gauss’s Law

The law states that flux out of a closed surface is equal to the charge enclosed divided by permittivity.

∮ E\overrightarrow{\rm E}.\overrightarrow{\rm dS}= 1/(\epsilon0).q

Gauss’s law states that : the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

It is easy to calculate electric field using Gauss’s law as compare to Coulomb’s law.(Gauss’s law is a replica of Coulomb’s law).

Steps to Find the Electric Field Using Gauss Law

Choose a Symmetrical Gaussian Surface: Select a closed surface where the electric field is constant in magnitude and direction. The symmetry can be spherical, cylindrical or planar depending upon the charge distribution.

• Determine the Enclosed Charge: Calculate the total charge enclosed in the chosen Gaussian surface.
• Calculate Electric Flux: Evaluate the flux of the electric field through the Gaussian surface ( use formula from fig5)
• Ultimately leads to finding the electric field using gauss law with the help of the point charge.

Let’s say you have a spherically symmetric charge distribution with total charge Q uniformly distributed within a sphere of radius R. Our aim is to find electric field at a distance r from the center:

• Choose a Gaussian Surface: A spherical Gaussian surface with radius r>R is suitable for this case.
• Determine Enclosed Charge: The enclosed charge is q for r<=R and Q for r>R (as whole body is enclosed).
• Calculate Electric Flux: The electric flux through the spherical surface is 4[Tex]\pi [/Tex]*r^2*E, where E is the magnitude of the electric field.
• Apply Gauss’s Law: By Set up Gauss’s law equation:

∮\overrightarrow{\rm E}.\overrightarrow{\rm dA} = qenc /\epsilon₀

4\pi.r² .E = Q/\epsilon₀

From this equation, you can solve for the magnitude of the electric field

E at a distance r from the center.

E = 1/4\pi.\epsilon₀ (Q/r² )

This example illustrate how to use Gauss’s law for a spherically symmetric charge distribution. See, the approach and the shape of the Gaussian surface can vary based on the symmetry of the charge distribution provided in the question.

Types of Charge Distribution

Let us talk about the different types of charge distribution:

• Linear Charge Distribution
• Surface Charge Distribution
• Volume Charge Distribution

Linear Charge Distribution

Linear charge distribution: If charge is distributed linearly over a body.

λ = dq/dl

λ = linear charge density

dq = charge

dl = line element

Surface Charge Distribution

Surface charge distribution: If charge is distributed continuously over the surface of a body

σ = dq/ds

σ = surface charge density

dq = charge

ds = surface element

Volume Charge Distribution

Volume charge distribution: If charge is distributed continuously over the volume of a body.

ρ = dq/dv

ρ = volume charge density

dq = charge

dv = volume element

Applications of Gauss Law to Find Electric Field

Now, let us see some of the application of gauss law to find electric field:

Electric Field due to a Line Charge

Suppose a line charge having linear charge density λ is given in the form of a thin charged rod.

Diagram for line charge

To find the electric field intensity at point P along a wire, a cylindrical Gaussian surface is selected. This choice is made to apply Gauss’s law for finding the electric field, E, at point P.

The electric flux passing through the end surfaces of the cylindrical Gaussian surface is ,that is Φ1 = 0.

And , the electric flux passing through the curved surface of the cylindrical Gaussian surface is given as:

Surface area of curved part is given as:

S = 2\pi*r*l

Total charged enclosed by gaussian surface is

q = \lambda*l

surface area and charge of gaussian surface (Equation-5)

The electric flux through the curved surface of the cylindrical Gaussian surface is given as:

Φ = [Tex]\overrightarrow{\rm E}.\overrightarrow{\rm dS} [/Tex]

= E. cos[Tex]\theta [/Tex].S

= E*1 *2 [Tex]\pi*r*l [/Tex]

flux through curved surface (Equation-6)

Total electric flux is given as:

Φ = Φ1 + Φ2

Φ = 0 + E.cos\theta.S

Φ = 2\pi*r*l*E

From Gauss’s law, we know that,

Φ = q/\epsilon0 = ????l/\epsilon0

\Longrightarrow 2\pi*r*l*E = \lambda l/\epsilon0

\Longrightarrow E = 1/2\pi\epsilon0 .(\lambda/r)

Electric Field Due to Ring

Now, let us take a look into the electric field due to

Electric Field Intensity at Any Point on the Axis of a Uniformly Charged Ring.

Diagram showing electric field due to charged ring

Let us consider a wire forming a circular ring with negligible thickness and a radius of R, carrying a uniform charge +q distributed evenly around its circumference. Our aim is to calculate the electric field intensity at any point P along the axis of the loop, positioned at a distance x from the ring’s center, marked as O.(as shown in figure-3)

Let AB be the length of element dl.

The charge on the element AB is,

dq =q*dl/2πR

Electric field intensity at P due to charge element AB is,

|dE| = K dq/(CP)2

where, K = constant = 14,

|dE| =K dq/(R2 + x2)

Now, resolve the electric field intensity dE into two rectangular components, that is

dE sinθ along the y-axis and dE cosθ along the x-axis.

And for diametrically opposite elements of the charged ring, the perpendicular components of the electric field intensity will nullify each other, resulting in,

∫dE sinθ = 0.

Whereas components along the axis of the charged ring will undergo integration. That is,

∫dE cosθ.

Hence, the resultant electric field intensity E at P is | E | = ∫dE cosθ

In △OPC ,cos[Tex]\theta [/Tex] = OP/CP = x/√(R² + X² )

Therefore,

|E| = ∫ k*x*d*a/(R² + X² )(√(R² + X² )

|E| = k*a*x/(R² + X² )^3/2

The direction of E is along the positive x-axis of the loop.

Electric Field Due to a Uniformly Charged Sphere

Now, let us talk about the electric field due to a uniformly charged sphere.

Charged sphere

Field seen Outside the Shell

To find out the electric field intensity at a point P outside the spherical shell when OP = r.

The Gaussian surface was taken as a sphere having radius r, while the electric field intensity will remain the same at every point seen on the Gaussian surface.

Thus, Gauss theorem becomes,

∮[Tex]\overrightarrow{\rm E}.\overrightarrow{\rm dS} [/Tex] = ∮ [Tex]\overrightarrow{\rm E}.\overrightarrow{\rm dS} [/Tex].n^ = q/[Tex]\epsilon [/Tex]₀

Or it can be also be given as,

E∮ dS = q/[Tex]\epsilon [/Tex]₀

Therefore, E.4.[Tex]\pi [/Tex].r² = q/[Tex]\epsilon [/Tex]₀

Therefore, the electric field become,

E = q/4[Tex]\pi.\epsilon [/Tex]₀ r²

From the above equation we can say that the electric field outside the shell is similar to the electric field due to a point charge. Thus outside the sphere, the electric field behaves as though it is due to a point charge (carrying all the charge of the shell) at the Centre of the shell.

Field Inside the Shell:

If the point P lies inside the spherical shell, then the Gaussian surface is a surface of a sphere having radius r. Since no charge is present inside the spherical shell, the Gaussian surface encloses no charge. Hence, q = 0.

???? = ∮ [Tex]\overrightarrow{\rm E}.\overrightarrow{\rm dA} [/Tex] = q_enc / [Tex]\epsilon [/Tex]₀

∮ EdAcos180 = q_enc/[Tex]\epsilon [/Tex]₀

-E.4[Tex]\pi [/Tex].r² = q_enc/ [Tex]\epsilon [/Tex]₀

-E.4.[Tex]\pi [/Tex].r² = 0/[Tex]\epsilon [/Tex]₀

E = 0

It proves that the field inside the spherical shell will always be zero.

Solved Examples on Electric Field

1. A force of 100 N is acting on the charge 10 μ C at any point. Determine the electric field intensity at that point.

Given:

Force F = 100 N

Charge q = 10 μ C

Electric field formula is given by

E = F / q

= 100N / 10×10−6C

E = 10^(7) N/C.

2. Calculate the electric field at points P, Q for the following two cases.(figure is provided below).

(a) For a charge of +1 µC placed at the origin.

Figure of part a

The magnitude of the electric field at point P is

E_p = 1/4.[Tex]\pi [/Tex].[Tex]\epsilon [/Tex]₀ (q/r² )

= (9 * 10⁹ * 1 * 10^-6 )/4

= 2.25 * 10³ NC^-1

Since, the source charges is positive, the electric field points away from the charge,

So the electric field at the point P is given by

[Tex]\overrightarrow{\rm E} [/Tex] = 2.25 * 10³ NC^-1 i^

For the point Q;

[Tex]\overrightarrow{\rm E} [/Tex]_Q = 9 * 10⁹ *1 * 10^-6 / 16 = 0.56 * 10³ NC^-1

Hence, [Tex]\overrightarrow{\rm E} [/Tex]_Q = 0.56 * 10³ j^.

(solution of part-a (Equation-11))

(b)For a charge of -2 µC placed at the origin

Figure for part -b

The magnitude of the electric field at point P is

E_p = 1/4[Tex]\pi.\epsilon [/Tex]₀ (q/r² )

= (9* 10⁹ * 2 * 10^-6)/4

= 4.5 * 10³ NC^-1

Since, the charge is negative, the electric field points towards,

So, the electric field at point P is given by

[Tex]\overrightarrow{\rm E} [/Tex] = -4.5 * 10³ NC^-1 i^

For the point Q :

( [Tex]\overrightarrow{\rm E} [/Tex]_Q = 9* 10⁹ * 1* 10^-6 )/36 = 0.5 * 10³ NC^-1

hence, [Tex]\overrightarrow{\rm EQ} [/Tex] = 0.5 * 10³ i^

(solution for part-b (Equation-12))

Application of Electric Field

• Electroporation: A technique in which electric fields are used to make pores in cell membranes to insert drugs, medicines, or genes. It is generally used in cloning processes.
• Electric fields play a role in studying tissue dynamics and controlling crystallization processes like nucleation and crystal growth, etc.

FAQs on Electric Field

1. What is an Electric Field?

An electric field at a particular location indicates the force exerted on a unit positive test charge placed at that point.

2. What is electric flux?

Electric flux represents the total number of electric field lines passing through a specified area per unit time.

3. What is a point charge?

A point charge refers to a theoretical charge situated at a point in space.