Lecture 2

Electric Field Lines

Electric field lines are an excellent way of visualizing electric fields. They were first introduced
by Michael Faraday himself. A field line is drawn tangentially to the net at a point. Thus at any
point, the tangent to the electric field line matches the direction of the electric field at that point.
Secondly, the relative density of field lines around a point corresponds to the relative strength
(magnitude) of the electric field at that point.
Electric Field Lines Attraction and Repulsion

Electric field lines always point away from a positive charge and towards a negative point. In
fact, electric fields originate at a positive charge and terminate at a negative charge.

As said before field lines are a great way to visualize electric fields. You can almost feel the
attraction between unlike charges and the repulsion between like charges as though they are
trying to push each other away.

What are the rules for drawing electric field lines?

Following are the rules for drawing electric field lines:

1. The field line begins at the charge and ends either at the charge or at infinity.
2. When the field is stronger, the field lines are closer to each other.
3. The number of field lines depends on the charge.
4. The field lines should never crossover.
Flux of an Electric Field
The total number of electric field lines passing a given area in a unit of time is defined as the
electric flux.

If the electric field is uniform, the electric flux passing through a surface of vector area A is:

ɸE = E.A

ɸE = EACosθ

where E is the magnitude of the electric field (having units of V/m), A is the area of the surface,
and θ is the angle between the electric field lines and the normal (perpendicular) to A.

For a non-uniform electric field, the electric flux dɸE through a small surface area dA is given
by dΦE=E⋅dA (the electric field, E, multiplied by the component of area perpendicular to the
field).

Gauss’s law

It states that the total electric flux through a closed surface (Gaussian Surface) is equal to the
charge enclosed divided by the permittivity.

Gauss’s Law Equation

As per the Gauss’s law, the total charge enclosed in a closed surface is proportional to the total
flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric
charge Q enclosed by the surface is

The Gauss law formula is expressed by

ϕ = Q/ϵ0

Where,

Q = Total charge within the given surface

ε0 = The electric constant

Let us now study Gauss’s law through an integral equation. Gauss’s law in integral form is given
below:

∫E⋅dA = Q/ε0

Where,

 E is the electric field vector

 Q is the enclosed electric charge

 ε0 is the electric permittivity of free space

 A is the outward pointing normal area vector

The Gaussian surface does not need to correspond to a real, physical object; indeed, it rarely will.
It is a mathematical construct that may be of any shape, provided that it is closed. However, since
our goal is to integrate the flux over it, we tend to choose shapes that are highly symmetrical. If
the charges are discrete point charges, then we just add them. If the charge is described by a
continuous distribution, then we need to integrate appropriately to find the total charge that resides
inside the enclosed volume.

Permittivity of free Space

The permittivity of free space, ε0, is a physical constant used often in electromagnetism. It
represents the capability of a vacuum to permit electric fields.
ϵ0 = 8.8542×10−12

Applications of Gauss’s law

Electric Field due to –ve Point Charge

The magnitude of the electric field E must be the same everywhere on a spherical Gaussian
surface concentric with the distribution. For a spherical surface of radius r:
 Φ=∮E.dA=E∮dA=E4πr2

According to Gauss’s law, the flux through a closed surface is equal to the total charge enclosed
within the closed surface divided by the permittivity of free space ϵ0. Let -q be the total charge
enclosed inside the distance r from the origin, which is the space inside the Gaussian spherical
surface of radius r. This gives the following relation for

Gauss’s law:
4πr2E=-q/ϵ0
-E4πr2 =-q/ϵ0
-ve sign, since E and A are opposite in direction

Hence, the electric field of a spherically symmetrical charge distribution has the following
magnitude and direction:

1 𝑞
E=
4πϵ0 𝑟2

Electric Field Due to Infinite Wire

Consider an infinitely long line of charge with the charge per unit length being λ. We can take
advantage of the cylindrical symmetry of this situation. By symmetry, the electric fields all point
radially away from the line of charge, and there is no component parallel to the line of charge.

We can use a cylinder (with an arbitrary radius (r) and length (l)) centered on the line of charge as
our Gaussian surface.

As you can see in the above diagram, the electric field is perpendicular to the curved surface of
the cylinder. Thus, the angle between the electric field and area vector is zero and cos θ = 1.
The top and bottom surfaces of the cylinder lie parallel to the electric field. Thus, the angle between
the area vector and the electric field is 90 degrees, and cos θ = 0.

Thus, the electric flux is only due to the curved surface.

According to Gauss law,

Φ = E.dA

Φ = Φcurved + Φtop + Φbottom

Φ = E . dA = ∫E . dA cos 0 + ∫E . dA cos 90° + ∫E . dA cos 90°

Φ = ∫E . dA × 1

Due to radial symmetry, the curved surface is equidistant from the line of charge, and the electric
field on the surface has a constant magnitude throughout.

Φ = ∫E . dA = E ∫dA = E . 2πrl

The net charge enclosed by the surface is

qnet = λ.l

Using Gauss theorem,

Φ = E × 2πrl = qnet/ε0 = λl/ε0

E × 2πrl = λl/ε0

E = λ/2πrε0

Electric Field due to Infinite Plate Sheet

Imagine an infinite plane sheet, with surface charge density σ and cross-sectional area A.
The position of the infinite plane sheet is given in the figure below:
The direction of the electric field due to infinite charge sheet will be perpendicular to the
plane of the sheet. Let’s consider cylindrical Gaussian surface, whose axis is normal to the
plane of the sheet. The electric field can be evaluated from Gauss’s Law as
 Infinite Charge Sheet
According to Gauss’s Law:

From continuous charge distribution charge q will be σ A. Talking about net electric flux,
we will consider electric flux only from the two ends of the assumed Gaussian surface.
This is because the curved surface area and an electric field are normal to each other,
thereby producing zero electric flux. So the net electric flux will be
Φ = 2EA
Then we can write

The term A cancel out which means electric field due to infinite plane sheet is
independent of cross section area A and equals to