Charge density formula refers to the amount of electric charge per unit volume or unit area in a given space. It is a measure of how much electric charge is distributed within a specified region. The charge density can be expressed in terms of volume charge density (charge per unit volume) or surface charge density (charge per unit area on a surface). Certainly! Charge density refers to the amount of electric charge per unit length, area, or volume. The specific formula for charge density depends on the geometric configuration of the charged object. Here are descriptions for linear charge density, surface charge density, and volumetric charge density: 1. **Linear Charge Density:** - Linear charge density (lambda) is defined as the amount of charge per unit length along a one-dimensional object, such as a wire or a rod. - The formula for linear charge density is given by: (lambda = {Q}/{L}), where (Q) is the total charge along the length (L). 2. Surface Charge Density: - Surface charge density (sigma) refers to the amount of charge per unit area on a two-dimensional surface. - The formula for surface charge density is: (sigma = {Q}/{A}), where (Q) is the total charge on the surface and (A) is the area of the surface. 3. **Volumetric Charge Density:** - Volumetric charge density (rho) represents the amount of charge per unit volume within a three-dimensional object. - The formula for volumetric charge density is: (rho = {Q}/{V}), where (Q) is the total charge enclosed within the volume (V). 4. **Disk Charge Density:** - For a charged disk, we often deal with surface charge density. The disk can be considered as a two-dimensional object, and the surface charge density (sigma) is defined as the charge per unit area. - The formula for surface charge density on a disk is: (sigma = {Q}/{pi R^2}), where (Q) is the total charge on the disk and \(R\) is the radius. 5. **Ring Charge Density:** - If you have a charged ring, the linear charge density (\(\lambda\)) is used. It is the charge per unit length along the circumference of the ring. - The formula for linear charge density on a ring is: \(\lambda = \frac{Q}{2\pi R}\), where \(Q\) is the total charge on the ring and \(R\) is the radius. These formulas help in quantifying how charge is distributed in various geometrical configurations, providing a useful framework for understanding and solving problems in electrostatics.

1. Charge density formula :
Charge density formula refers to the amount of electric charge per unit
volume or unit area in a given space. It is a measure of how much electric
charge is distributed within a specified region. The charge density can be
expressed in terms of volume charge density (charge per unit volume) or
surface charge density (charge per unit area on a surface).
.
Table of content (charge density formula and
concepts)
a). Discrete charge distribution
b). Continuous charge distribution
1.1 linear charge density
Linear charge distribution

2. 1.2 Analysis of electric field
1.2(a) Electric field due to uniform
charge Rod
1.2(b) Let's try some difficult problem
with it
1.2(c) let's try in finite case
1.2(d) electric field due to arc
1.2(e) some special cases
1.3 surface charge distribution
or surface charge density
1.3(a) electric field due to disc
1.3(b) Field due to annular disc
1.4 volume charge density
1.5 next topic will be 👍
Before diving into what is charge density and where it can be used, first we
have to know how many types of basic charge density.
So there Three three basic type of charge density.
1. linear charge density
2. surface charge density
3. volume charge density
and these are 3 types of charge densities become in picture by charge
distribution so if we see the charge distribution, how many types of charge
distribution then we will find out that there are two types of charge
distribution .

3. a). Discrete charge distribution
b). Continuous charge distribution
a). Discrete charge distribution
In easy language
When a charge is concentrated at a specific location this type of charge
distribution is called discrete charge distribution.
in other word
Discrete charge distribution in electromagnetism refers to the unique
scenario where electric charge is segregated into distinct and separate
points, as opposed to being continuously dispersed over an area. In
simpler terms, the electric charge is confined to specific points or locations
within a defined space, each linked to a specific quantity of electric charge.
This particular charge distribution pattern is commonly encountered in
situations where electric charges are concentrated at individual particles
like electrons or protons. In contrast, continuous charge distribution
involves the uniform spread of charge across a designated region.

4. In the mathematical realm of discrete charge distribution, the total charge
(Q) is calculated by summing up the electric charges (q_i) at each discrete
point:
[ Q = Σi q_i ]
Here, (q_i) denotes the electric charge at each specific location.
Understanding discrete charge distribution is imperative for addressing
challenges in electrostatics and magnetostatics, as the arrangement of
charges significantly influences the electric and magnetic fields within a
given system.
b). Continuous charge distribution
In easy language
When charges are distributed over the entire body, not only specific points,
this type of charge distribution is called continuous charge distribution.
some point related to it
● in this case infinite number of charges are closely packed
● and distribution is uninterrupted and continuous

5. ● no space between them unlike discrete charge distribution
Example 👍
like charged sphere, charged rod, charged disc and so on.
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Now we are close to the distribution type of charges so let's get started on
how many types of the charge distribution as we discuss above there are
three types so let's get started one by one.
1.1 linear charge density
Or linear Charge Density:
In Easy language
when charge is linearly distributed over length of a body this type of
distribution is called linear charge reservation
and in other word
Electric charge distribution along a one-dimensional entity, such as a wire
or line, is encapsulated by the concept of linear charge density. This
parameter delineates the amount of electric charge present per unit length

6. along the linear object. In mathematical terms, linear charge density is
symbolized by λ (lambda) and can be expressed as follows:
charge density formula for it.
[ λ = limΔl->0 (ΔQ/Δl)]
In this equation, (ΔQ ) represents the charge contained within a small
length (Δl ) and the limit is calculated as ( Δl) approaches zero. Essentially,
linear charge density signifies the charge per unit length in the scenario
where the length becomes infinitely small.
1.2 analysis of electric fields.
1.2(a) Field due to uniform charge rod
At axial Position
as we take a small element of this rod dx which is contain dq (small
charge) and Totle charge contained by rod is λ.

7. So this is an expression for the axial electric field. By the same method we
can get a perpendicular electric field.
As we get the formula of electric field due to uniform charge rod we have
got two components of electric field first , electric field in the parallel
direction of rod and electric field in the vertical direction of rod.
E丄 = (k λ／r) *(sinα + sinβ) vertical component
E|| = (k λ／r) *(cosα - cosβ) horizontal component

8. Where
r :-> perpents distance between rod and point
α (Alpha) :-> angle between perpendicular and line of joining to the upper
end
β(Beta):-> is angle between perpendicular and line of joining to the lower
end
Note :-> α (Alpha) and β(Beta) are measured in opposite directions.
Let's know how can be use linear charge density
● Q. First we will take a positively charged rod and try to find out the
electric field due to it At a distance d so let's get started.
we assume that charge rod charged with λ charge density and we
will find out electric field at a point p and angle between upper end of
rod and perpendicular distance to the point , 30 degree and angle
between lower end of rod and perpendicular distance between point
is 30 degree so let's start ..

9. As we know the electric field due to this rod is going to be two parts:
first one perpendicular electric field and second one parallel electric
field so let's find out one by one .
Perpendicular electric field due to this rod
E丄 = (k λ／r) *(sinα + sinβ) …..(1)
Parallel electric field due to this rod
E|| = (k λ／r) *(cosα - cosβ) …..(2)
α = 30, β = 30 ……(given)
So let's put the values into equation number one
E丄 = (k λ／r) *(sinα + sinβ)
E丄 = (k λ／r) *(sin30 + sin30)
E丄 = (k λ／r) *(1/2 +1/2)
E丄 = (k λ／r) so this is E.Field in perpendicular direction.
Similarly for parallel direction.
E|| = (k λ／r) *(cosα - cosβ)

10. E|| = (k λ／r) *(cos30 - cos30)
E|| = 0 so this is E.Field in parallel direction.
● Q.
Now let's try the same question but from a different angle .
Now we have considered the angle between upper end and
perpendicular distance, 37 degree and Angle between lower end
perpendicular distance between that point is 52 degree now let's find
out Electric field at one P.
As we know the formula of electric field due to charged rod
Perpendicular electric field due to this rod
E丄 = (k λ／r) *(sinα + sinβ) …..(1)
Parallel electric field due to this rod
E|| = (k λ／r) *(cosα - cosβ) …..(2)
So let's put the values of angles into these formulas and find out electric
field

11. E丄 = (k λ／r) *(sin37 + sin52)
[sin37 = 3/5 , sin52 = 4/5] by triangular pairs
E丄 = (k λ／r) *(3/5 + 4/5)
E丄 = (k λ／r) *(7/2)
Similarly
E|| = (k λ／r) *(4/5 - 3/5)
E|| = (k λ／r) *(1/5)
● Q.
This time we will take a negative rod and let's find out what is the
electric field at point P.
as we know that electric field due to negative charge is towards the
body as we read so E.field in this case towards the rod so it's
interesting to find out electric field.
as we know electric field due to Rod
E丄 = (k λ／r) *(sinα + sinβ)
E|| = (k λ／r) *(cosα - cosβ)

12. In this case only perpendicular electric fields exist due to parallel
electric fields being canceled to each other so only the perpendicular
component of the electric field remains .
( I'm assuming that you are very well to know how to take
component of electric field or electric forces. )
So
E丄 = (k λ／r) *(7/2) (-j)
So this one is the same as the above example but direction of electric
field is negative and it is obvious due to being known that negative
electric fields towards the object and positive electric field outwards
so in this case electric field towards the rod not away from the rod.
1.2(b) Let's try some difficult problem problems with it
● Q. What will be the electric field at point P if angle between upper
end and perpendicular distance is 45 degree.

13. As we can see in the figure that we have only given one angle α and
by seeing the figure the value of β is zero. so
α = 45 , β = 0
let's put the values and find out.
E丄 = (k λ／r) *(sinα + sinβ)
E丄 = (k λ／r) *(sin45 + sin0)
E丄 = (k λ／r) *( 1/2 + 0 )
E丄 = (k λ／r)/2
And
E|| = (k λ／r) *(cos45 - cos0)
E|| = (k λ／r) *(1/2 - 1) (-ve)
In this case when we calculate, we will find that the electric field in
the parallel direction will be negative so it means the direction of the
parallel electric field is not upward so it is downward.

14. 1.2(c) Let's try infinite case
Now we are going to introduce an infinite case due to the charged rod. It is
nothing but when we take a point P and put it very near to the charged rod
then the electric field at that point is max so this is the case which we call
an electric field due to the infinite rod.
First we put point P some distance to the rod and we can observe that
angle between upper end of rod and perpendicular distance α and angle
between lower end of rod and perpendicular distance β.
When we decrease distance between P point and rod then electric field will
increase and when the point P becomes very close to the rod then electric
field will be max and angle Exerted by upper in and lower end is 90 degree.
note : So in respect of point P the charged rod behaves like an infinite wire.
E丄 = (k λ／r) *(sin90 + sin90)
E丄 = 2*(k λ／r) (max)
E|| = (k λ／r) *(cos90 - cos90)

15. E|| = (k λ／r) * 0
E|| = 0
and electric field due to infinite wire [E = 2*kλ/r ] (radially outward)
1.2(d) Electric field due to arc
Now we are going to see what electric field is due to arc so in this case we
have taken an Arc and we provide charge λ and we have to find out electric
field at point c.
so when we take the electric field due to the most left part of Arc and
electric field due to the most right part of arc and after finding the
resultant electric field.
The direction of the resultant electric field along the angle bisector of that
arc.
As we read already that when two equal forces we have given and have to
find out resultant force then we use 2Fcos(θ/2) so this is the same case
but only difference is 2Esin(θ/2).

16. So Electric field due to arc
E = 2Esin(θ/2) = 2* k λ/r *sin(θ/2)
similarly :
1.2(e) Some special cases
Q. Electric field due to semi infinite war
In this case we will take an infinite charge rod or wire and try to find out
what the electric field is at point P.
So in this case we will use the same formula which we have read in case of
charge Rod, parallel and perpendicular electric field .

17. E丄 = (k λ／r) *(sinα + sinβ)
E|| = (k λ／r) *(cosα - cosβ)
As we can see in this figure the value of (α) Alpha ( angle between upper
end of wire to the perpendicular distance) is like 90 degree due to infinite
wire and value of (β) Beta is 0.
So
E丄 = (k λ／r) *(sin90 + sin0)
E丄 = (k λ／r)
And
E|| = (k λ／r) *(cos90 - cos0)
E|| = (k λ／r) *(- 1)
E|| = - (k λ／r)
So the direction of the parallel electric field is downward not upward due
to negative sign.
Q. Electric field due to horse shoe
In this case we will use two concepts: first one electric field due to semi
infinite wire and second one electric field due to semicircle so let's find out
the electric field at point P.

18. As we already read above electric field due to semi-finite war is
E丄 = (k λ／r)
E|| = - (k λ／r)
and electric field due to semicircle
E = 2(k λ／r)
as we can see in the figure electric field (E丄) due to Rod Is canceled by
electric field (E丄) due to another Rod And also we can see electric field
due to semicircle is canceled by parallel component of electric field due to
both rods.
So overall the electric field at the point P is zero.
The unit of measurement for linear charge density is commonly articulated
in coulombs per meter (C/m) within the International System of Units
(SI).

19. This parameter frequently comes into play when dealing with problems
related to charged wires, cables, or other elongated conductive objects
where the charge is distributed along their length. In such cases,
understanding and manipulating linear charge density become pivotal for
effective analysis and problem-solving within the realm of electric
phenomena.
1.3 Surface Charge Distribution:
Or Surface Charge Density:
Easy language:
Surface charge density , when charges are distributed over the surface of
the body , is called surface charge density or surface charge distribution.
Other word
In this configuration, electric charge gracefully extends over a
two-dimensional surface. The surface charge density ( 𝝈) takes center
stage, representing the charge per unit area on the surface, with the
flexibility to fluctuate across the surface.
The mathematical formulation for surface charge density ( 𝝈) is articulated
as:
charge density formula for it.
[ 𝝈 = limΔA->0 (ΔQ/ΔA) ]
In this context, (ΔQ) embodies the charge residing on a small area (ΔA).
Let us know how can we use this surface density:

20. 1.3(a) Electric field due to disc
When we provide charge to a disc then all charges are distributed over the
surface of it, meaning over the area of that disc so let's find out what is the
electric field due to it ?
In case of disc or surface charge density we will use surface charge density
(𝝈) so
electric field at point P due to disc is.
E = (𝝈/2ε)*(1-cosθ)
As we can see in the figure when we increase angle between upper part of
disc and disc bisector line then cos theta will reduce.
as we know that while increasing θ, cosθ will decrease.
Note:->
● while moving towards the Origin of X-axis electric field will be
increased due to (cosθ ) will be increased.

21. ● We know [ cos30 > Cos60] so when We increase the value of (θ)
theta then the electric field also increases due to (cosθ ) will reduce .
● When (cosθ ) becomes zero (0) then the electric field due to the disc
will be maximum and this case is called the infinite disc case or
infinite seat.
E = (𝝈/2ε)*(1-cos90)
E = 𝝈/2ε
1.3(b) Electric field due to annular disc
Electric field due to the annular disc is the same as the disc but in this case
we will use two angle alpha and beta so electric field at point P.
E = (𝝈/2ε) (cosα -cosβ)

22. The prevalence of continuous charge distributions transcends various
physical systems, finding applications in charged conductors, dielectric
materials, or regions hosting a net charge dispersed across either volume
or surface. A profound grasp of continuous charge distributions proves
fundamental in the study of electromagnetism, playing a pivotal role in
unraveling complexities associated with electric fields, electric potential,
and various electrostatic phenomena.
1.4 Volume Charge Distribution:
This scenario unfolds as electric charge permeates a three-dimensional
volume. The charge density (𝝆) comes into play, denoting the charge per
unit volume, dynamically altering its manifestation based on the position
within the volume.
Expressing this mathematically, we define (𝝆) as:
charge density formula for it.
[ (𝝆) = limΔV->0 (ΔQ/ΔV) ]
Here, ΔQ signifies the charge contained within a minute volume ΔV.
1.5 Next topic will be 👍
● Electric flux
● concept of projected area concept of projected area with electric flux
● flux association for cube
● Difference between surface and close surface

23. ● flux in Variable electric field
● Electric field due to ring
And so on …..