Electric Field due to Continuous Charge Distribution - Physics Class 12

Introduction

With the help of Coulomb's law and the principle of superposition, the electric field due to a system of discrete charges can be found by adding vector contributions of each point charge. In the limit in which the individual charges become infinitesimally small and closely packed, the charge distribution becomes continuous and calculus (integration) is used to find the resultant field.

Consider a system of discrete charges q₁, q₂, q₃, ..., q_n. The net charge is obtained by algebraic sum of these charges and the net electric field at a point is obtained by vector addition of fields due to each charge using superposition. When the number of charges becomes very large and they are distributed over a length, surface or volume, we treat the distribution as continuous and replace summation by integration (Δq → dq).

• Discrete charge systems involve a finite number of point charges, whereas continuous charge systems involve an extremely large number of charges distributed over a length, surface or volume.
• Continuous charge distributions require calculus and are used to model line, surface and volume charge systems.

When the size of individual charges becomes very small and their number becomes extremely large, summation is replaced by integration.

Considering the charge distribution as continuous, the total field at a point P is obtained in the limit Δq_i → 0 by integrating the infinitesimal field contributions dE from all dq elements distributed over the source. 

This combination of infinitely many point charges distributed over a line, surface or volume constitutes respectively a linear, surface or volumetric charge distribution, with corresponding charge densities λ, σ and ρ.

Types of Continuous Charge Distribution

1. Linear charge distribution

A body having charge distributed along one dimension (length) has a linear charge distribution. The linear charge density is denoted by the Greek letter λ (lambda) and is defined so that the charge on an infinitesimal length element dl is dq = λ dl.

Observe the rod given above of length L, a charge of +Q is distributed along the length of the rod. A small element dl will have a charge dq on itself. In this case, we define linear charge density of the rod.

dq = λdl [Charge on infinitely small element dl]
Q = ∫dq = ∫λdl [Total charge on the rod]

2. Surface charge distribution

A body having charge distributed over a two-dimensional surface has a surface charge distribution. The surface charge density is denoted by σ (sigma). For a small surface element dA, the infinitesimal charge is dq = σ dA and the total charge is Q = ∫ σ dA.

dq = σdA [Charge on infinitely small element d A]
Q = ∫dq = ∫σdA [Total charge on the sheet]

3. Volume (volumetric) charge distribution

A body having charge distributed throughout a three-dimensional region has a volume charge distribution. The volume charge density is denoted by ρ (rho). For a small volume element dV, the infinitesimal charge is dq = ρ dV and total charge is Q = ∫ ρ dV.

dq = ρdV [Charge on infinitely small volume element dV]
Q = ∫dq = ∫ρdV [Total charge on the body]

In NEET problems, the choice of infinitesimal element (dl, dA or dV) is crucial and depends on the symmetry of the charge distribution.

Charge Densities - Definitions and Units

Linear charge density

When charge is distributed along the length of a body, it is called linear charge distribution. The distribution may be uniform or non-uniform. It is also called linear charge density and is denoted by the symbol λ (Lambda).

The linear charge density is defined by

λ = dq/dl

Unit: coulomb per metre (C m⁻¹).

If we consider a conductor of length ‘L’ with surface charge density λ and take an element dl on it, then small charge on it will be
dq = λ dl

Fig: We take a small element x and integrate it in case of linear charge density

Surface charge density

When charge is distributed over the surface of a body, it is called surface charge distribution. The surface charge density may be uniform or non-uniform. It is denoted by the symbol σ (sigma) symbol.

The surface charge density is defined by

σ = dq/dA where dq is the small charge element over the small surface dA.

Unit: coulomb per square metre (C m⁻²).

For a small surface element dA, dq = σ dA.

Volume charge density

When the charge is distributed over a volume of the body (generally an insulator), it is called Volume Charge Distribution. It is denoted by symbol ρ (rho). In other words charge per unit volume is called Volume Charge Density.

The volume charge density is defined by

ρ = dq/dV where dq is small charge element located in small volume dv.

Unit: coulomb per cubic metre (C m⁻³).

For a small volume element dV, dq = ρ dV.

Fig: We can easily find electric field in different geometries using charge distribution system

Charge density may be uniform (constant) or non-uniform (a function of position), depending on the physical situation.

General Procedure to Calculate Electric Field from a Continuous Distribution

(i)Identify the type of charge distribution and compute the charge density λ, σ or ρ.

(ii) Divide the charge distribution into infinitesimal charges dq, each of which will act as a tiny point charge.

(iii) The amount of charge dq, i.e., within a small element dl, dA or dV is
dq = λ dl (charge distributed in length)
dq = σ dA (charge distributed over a surface)
dq = ρ dV (charge distributed throughout a volume)

(iv) Draw at point P the dE vector produced by the charge dq. The magnitude of dE is

(v) Resolve the dE vector into its components. Identify any special symmetry features to show whether any component(s) of the field that are not canceled by other components.

(vi) Write the distance r and any trigonometric factors in terms of given coordinates and parameters.

(vii) The electric field is obtained by summing over all the infinitesimal contributions.

(viii) Perform the indicated integration over limit of integration that includes all the source charges.

Electric Field - Selected Standard Results (Uniform Distributions)

(a) Electric field on the axis of a uniformly charged circular ring

Consider a uniformly charged circular ring with a total charge +Q distributed uniformly along its length. We need to evaluate the net electric field due to this charged ring at a point P which is located x distance from its centre on its axis.
Conclude that the charge is distributed linearly throughout the length of the ring, hence we will define linear charge density λ for this ring,
λ = Total charge on the ring/𝑇otal Length =Q/2πr
Now, we consider an infinitely small length element dl on the ring,
Infinitesimal charge on element dl,
dq = λdl
Now, we write the expression of Electric Field at point P due to dq

This, infinitely small electric field vector will be inclined at an angle θ with the axis of the ring (x axis), as shown in diagram.
We need to imagine components of  along the x and y axis i.e. and
By resolvingwe get,


Observe and imagine, thatwill cancel out if we take each and every element of the ring into consideration.
Therefore net electric field at P,

Now, by geometry,

Thus,



Replacing with the value of λ defined above,

This is the standard expression for the field on the axis of a uniformly charged ring: the field points along the axis and vanishes at the centre (x = 0) for a uniformly charged ring. This happens due to complete symmetry of charge distribution about the centre of the ring.

(b) Electric field at a general point due to a uniformly charged finite rod

Consider a straight rod with uniform linear charge density λ. Let the perpendicular distance from the rod to the field point P be r. Choose a coordinate x along the rod and take an element of length dx at position x measured from a chosen origin on the rod.

Now if dE be the electric field at P due to the element, then it can be given as

Here

Now we resolve electric field in components. Electric field strength in x-direction due to dq at P is,
dE_x = dEsin 𝜃


Here we have x = r tan 𝜃
and dx = rsec² 𝜃d𝜃

Net electric field strength due to dq at point P in x-direction is

or

Similarly, the electric field strength at point P due to dq in y-direction is
dE_y = dEcos𝜃

Again we have x = rtan 𝜃
And dx = rsec² 𝜃d𝜃
Thus we have,

Net electric field strength at P due to dq in y-direction is



Thus electric field at a general point in the surrounding of a uniformly charged rod which subtends angles 𝜽₁ and 𝜽₂ at the two corners of the rod from the point of consideration can be given as In parallel direction,

In perpendicular direction ,

Here θ₁, θ₂ are the angles made by the lines joining P to the ends of the rod with the perpendicular from P to the rod (so that tan θ = x/r). These formulae are convenient because many problems give angles subtended at the point.

In NEET, problems often directly provide θ₁ and θ₂, making these expressions especially useful.

Special cases obtained from the finite expression:

(i) Infinitely long uniformly charged rod with charge density 𝜆:

For infinite rod, 𝜃₁ → 90° and 𝜃₂ → 90°
Therefore, for infinitely long uniformly charged rod,

While,

(ii) Electric field due to semi-infinite wire:
For this case,

(c) Electric field on the axis of a uniformly charged disc

If there is a disc of radius R, charged on its surface with surface charge density s C/m², we wish to find electric field strength due to this disc at a distance x from the centre of disc on its axis at point P shown in figure.

Note: Identify that the electric charge is distributed over the surface of the non-conducting disc, hence we would define a surface charge density σ for this disc.
σ = Total Charge/Total Area = Q/πR²

To find electric field at point P due to this disc, we consider an elemental ring of radius y and width dy in the disc as shown in figure. Now the charge on this elemental ring dq can be given as
dq = σ (dA)
where dA is the area of the ring element on the disc,
also we can imagine ring element to be a small rectangle with width dy. Thus,
dA = 2πydy
dq = σ(2πydy)
Now we know that electric field strength due to a ring of radius R. Charge Q at a distance x from its centre on its axis can be given as

Here due to the elemental ring electric field strength dE at point P can be given as

Net electric field at point P due to this disc is given by integrating above expression from O to R as


Now, using integration by substitution we can solve the above integral as,


By geometry,

Hence,

Please note that 𝜽 is the angle subtended by the disc at point P which is x distance far from the center.

Two important limiting cases:

Case: (i) If x < < R ⇒ θ → 90° ⇒ cosθ → 0 Physically, this would mean that the disc has its radius R→ ∞, that is the disc can be effectively imagined as infinitely long sheet of charge,Thus, Electric field due to infinitely long plane sheet of charge at a distance x would be,i.e. behaviour of the disc is like infinite sheet.

Case: (ii) If x > > R

Now, using binomial approximation,

i.e. behaviour of the disc is like a point charge.

Electric field due to a uniformly charged hollow hemispherical shell (thin shell)

Figure shows a hollow hemisphere, uniformly charged with surface charge density 𝜎 C/m² . To find electric field strength at its centre C, we consider an elemental ring on its surface of angular width dθ at an angle θ from its axis as shown. 

The surface area of this ring will be

dA = 2𝜋r × Rdθ
By geometry, dA = 2𝜋R sin 𝜃 × Rdθ
Charge on this elemental ring is
dq = 𝜎d𝐴 = 𝜎. 2𝜋R sin 𝜃 × Rdθ

Now due to this ring electric field strength at centre C can be given as,

Net electric field at centre can be obtained by integrating this expression between limits 0 to 𝜋/2.

Hence, Electric Field intensity at centre C, due to uniformly charged nonconducting hemispherical shell is,

The electric field is directed along the symmetry axis, away from the curved surface.

Above given continuous charged systems are most frequently used ones. It is recommended to remember the procedure and results by heart.

Remarks and Important Formulae (for quick reference)

• Linear density: λ = dq/dl. Surface density: σ = dq/dA. Volume density: ρ = dq/dV.
• Field on axis of ring of radius R, total charge Q, at distance x from centre:
  E = (1/4πε₀) · Q x / (R² + x²)^(3/2).
• Field on axis of a uniformly charged disc (radius R, density σ) at distance x:
  E = (σ / 2ε₀) · (1 - x / √(x² + R²)).
• Field of an infinite plane sheet: E = σ / (2ε₀) (independent of distance).
• Field of an infinitely long line of charge: E = λ / (2πε₀ r) directed radially outward (r = perpendicular distance).
• Field at centre of thin hemispherical shell (uniform σ): E = σ / (4ε₀) along the axis.

Worked-out approach checklist (to use for any continuous distribution)

• Choose coordinates and an appropriate infinitesimal element (dl, dA, dV) that exploits the symmetry of the problem.
• Write dq in terms of the chosen density and element.
• Write dE using Coulomb's law and resolve into components; use symmetry to drop cancelled components.
• Express geometry (distance, cos θ or sin θ) in terms of the integration variable.
• Integrate with correct limits and simplify; check limiting cases (near field and far field) to verify results.

Summary: Continuous charge distributions are modelled by linear density λ, surface density σ or volume density ρ. The field at any point is obtained by integrating infinitesimal Coulomb contributions dE = (1/4πε₀) (dq / r²) r̂. Symmetry simplifies the calculation by cancelling components; standard results for ring, rod, disc, plane and hemisphere are useful to memorise and apply. These standard results are frequently used directly in NEET multiple-choice questions.