## 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).

.

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.

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 .

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.

**linear charge density****surface charge density****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 .

**a). Discrete charge distribution**

b). Continuous 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.

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
- no space between them unlike discrete charge distribution

Example 👍

like charged sphere, charged rod, charged disc and so on.

** **

**Learn more excitement topics:**

**learn more :**

- electric field and application (in easy way )
- electric force most important concepts with examples
- charge density (linear, area, volume with most helpful important examples )
- in-depth about charge with experiments
- what is electric flux: know every things about it with iits examples
- Application of gauss law

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 Farmula:

**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 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.**

E丄 = (k λ／r) *(sinα + sinβ) vertical component

E|| = (k λ／r) *(cosα - cosβ) horizontal component

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 ..

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)**

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.

**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

**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)

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)

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)In this case only perpendicular component of electric fields exist due to parallel electric fields being canceled to each other so only the perpendicular component of the electric field remains .

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

**(-j)**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.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.

**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.

**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)

E|| = (k λ／r) * 0

E|| = 0

E丄 = (k λ／r) *(sin90 + sin90)

E丄 = 2*(k λ／r) (max)

E|| = (k λ／r) *(cos90 - cos90)

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).**

**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.

**E丄 = (k λ／r) *(sinα + sinβ)**

E|| = (k λ／r) *(cosα - cosβ)

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)

E丄 = (k λ／r)

**And**

**E|| = (k λ／r) *(cos90 - cos0)**

E|| = (k λ／r) *(- 1)

E|| = - (k λ／r)

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

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.

**E丄 = (k λ／r)**

E|| = - (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).

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 farmula:

**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:**

**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 ?**electric field at point P due to disc is.**

**E = (𝝈/2**

**ε)*(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.

- 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/2**

**ε)*(1-cos90)**

E = 𝝈/2

E = 𝝈/2

**ε**

**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)**

**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 👍

And so on …..**learn more :**

- electric field and application (in easy way )
- electric force most important concepts with examples
- charge density (linear, area, volume with most helpful important examples )
- in-depth about charge with experiments
- what is electric flux: know every things about it with iits examples
- Application of gauss law