The continuous load distribution system is a system in which the charge is uniformly distributed over the conductor. For a continuous charging device, the infinite number of charges is closely packed and there is no space between them. Unlike the discrete charging system, the continuous load distribution in the conductor is uninterrupted and continuous. There are 3 types of continuous charge distribution system -

Linear Charge Distribution

Surface Charge Distribution

Volume Charge Distribution

If the charge is not evenly distributed over the length of the conductor, it is called linear charge distribution. It is often referred to as linear charge density and is denoted by the Lambda (λ ) symbol. Mathematically, there is a linear charge density -

λ = dq/ dl

The unit of the linear load density is C / m. If we find a conductor with a length of 'L' with a surface load density of λ and take an aspect of dl on it, then a small charge will be on it.

dq = λ dl

So, there will be an electrical field on the small charge element dq.

dE = kdq/r2

dE = kλdl / /r2

In order to calculate the net electrical field, we will integrate both sides with the correct limit, i.e.

\[\int\] dE = \[\int_{0}^{L}\] \[\frac{k\lambda dl}{r^{2}}\]

\[\int\] dE = \[\frac{k}{r^{2}}\] \[\int_{0}^{L}\] \[\lambda\] dl

Surface Charge Density

When the charge is uniformly distributed over the conductor surface, it is called Surface Charge Density or Surface Charge Distribution. It is denoted by the symbol σ (sigma) and the unit is C / m2. It is also defined as a charge/per area of the unit. Mathematically the density of the surface charge is

σ = dq / ds

Where dq is a small element of charge over a small ds surface. So, there will be a small charge on the driver -

dq = σ ds

The electric field due to a small charge at a distance of 'r' can be evaluated as

dE = kdq/r2

dE = kσdl / /r2

Integrating both sides with proper limits we get

\[\int\] dE = \[\int_{0}^{s}\] \[\frac{k \sigma ds}{r^{2}}\]

\[\int\] dE = \[\frac{k}{r^{2}}\] \[\int_{0}^{s}\] \[\sigma\] ds

When the charge is distributed over a driver's volume, it is called Volume Charge Distribution. It is denoted by the ρ (rho) symbol. In other words, the charge per unit volume is called Volume Charge Density and its unit is called C / m3. Mathematically, the density of the volume charge is

ρ = dq/ dv

where dq is a small charge element located in a small volume dv. To find the total charge, we must integrate dq with the correct limits. The electric field due to dq will be

dq = ρ dv

dE = kdq / r2

dE = kρdv / /r2

Integrating both sides with proper limits we get

Calculation of Electric Field

Let us consider the case of the continuous distribution of charges in the body. Here, because of this charge, we'll measure the electrical field at point P. We can say that the charge density of different volumetric elements can be different so that we divide the body into different elements so that the charge density of a particular element can be considered to be a fixed quantity. Consider one of those elements of volume Δv, the charge density of which is given by ρ. Let the distance between the volume element and point P be given as r. The charge in the volume element could be given as ρΔv. As per the law of Coulomb, the electrical field can be given as ρΔv due to the charge.

\[\Delta\]E = \[\frac{1}{4\pi \epsilon_{0}}\] \[\frac{p 𝚫 V}{r^{2}}\] \[\overbrace{r}\]

Here, r is the distance between the charged part and the point P at which the field is to be measured, and ř is the unit vector in the direction of the electrical field from the charge to point P. By the principle of superposition, the electrical field can be given as a result of the total distribution of the charge in the body, which is divided into a number of such volume elements.

E = \[\frac{1}{4\pi \epsilon_{0}}\] \[\sum_{all 𝚫 V}^{}\] \[\frac{p𝚫V}{r^{2}}\] \[\overbrace{r}\]

Electrical field due to the above volume charge distribution (for 1 and n volume elements) will be as follows -

\[\Delta\]E = \[\frac{1}{4\pi \epsilon _{0}}\] \[\frac{p𝚫V}{r^{2}}\] \[\widehat{{r}'}\] \[\Delta\]E \[\approx\] \[\frac{1}{4\pi \epsilon _{0}}\] \[\sum\] \[\frac{p𝚫V}{r^{2}}\] \[\widehat{{r}'}\]

FAQ (Frequently Asked Questions)

Q1 - A rod of length l with uniform charge per unit length λ is placed at a distance d from the origin along the x-axis. A similar rod is placed at the same distance along the Y-axis. Determine the magnitude of net electric field intensity at the origin.

Solution -

Let's first find Electric field due to the rod on x-axis, say E1

Consider a small length element dx, distance x away from origin.

Field due to this dE1 = kλdx / x2

Integrating from x = d to x = d+l

Gives E1= λl / 4πϵ0d(d+l) along -x direction

Now, E2is along the -y direction and is the same in magnitude.

Thus,

∣E∣ = √2∣E1∣ = √2λl / 4πϵ0d(d+l)

Q2 - What do you mean by continuous charge distribution. Also define uniform and non-uniform charge distribution?

Ans - In a continuous distribution of charges, all charges are closely linked together, i.e. they have much less space between them. But this tightly bound system does not mean that the electrical charge is uninterrupted. It is obvious that the distribution of separate charges is continuous, with little space between them.

Uniform Charge Distribution

If the charge is not concentrated in any section but distributed uniformly, then it is considered a uniform distribution of the charge.

Non-uniform Charge Distribution

If the charge is not uniformly spread over the surface, the distribution of the charges is not uniform.