Electric Field due to Continuous Charge Distribution JEE Notes | EduRev

Physics Class 12

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ELECTRIC FIELD DUE TO CONTINUOUS CHARGE DISTRIBUTION
With the help of Coulomb’s Law and Superposition Principle, we can easily find out the electric field due to the system of charges or discrete system of charges. The word discrete means every charge is different and has the existence of its own. Suppose, a system of charges having charges as q1, q2, q3……. up to qn. We can easily find out the net charge by adding charges algebraically and net electric field by using the principle of superposition.

This is because:

  • Discrete system of charges is easier to solve
  • Discrete system of charges do not involve calculus in calculations

Electric Field due to Continuous Charge Distribution JEE Notes | EduRevElectric Field due to Continuous Charge Distribution JEE Notes | EduRevConsidering the charge distribution as continuous, the total field at P in the limit Δqi → 0 is
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
This means a combination of infinite point charges kept together forming a linear, surface or a volumetric shape constitutes a continuous charge system with linear, surface or volumetric charge density respectively.

Refer to the following figure:
Electric Field due to Continuous Charge Distribution JEE Notes | EduRevThus, there are three types of continuous charge distribution system.
(i) Linear Charge Distribution: A body having a finite charge distributed along its length i.e. along one dimension will have a linear charge distribution. In this case, we define the Linear Charge Distribution denoted by lower case Greek letter lambda (λ).
Electric Field due to Continuous Charge Distribution JEE Notes | EduRevObserve 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.
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev

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

(ii) Surface Charge Distribution: a body having a finite charge distributed along its area or surface will have a Surface Charge Distribution. In this case, we define the Surface Charge Distribution denoted by lower case Greek letter Sigma (σ).
Electric Field due to Continuous Charge Distribution JEE Notes | EduRevElectric Field due to Continuous Charge Distribution JEE Notes | EduRev
dq = σdA [Charge on infinitely small element d A]
Q = ∫dq = ∫σdA [Total charge on the sheet]

(iii) Volume Charge Distribution: a body having a finite charge distributed along its volume will have a Volumetric Charge Distribution. In this case, we define the Volumetric Charge Distribution denoted by lower case Greek letter rho (ρ).
Electric Field due to Continuous Charge Distribution JEE Notes | EduRevElectric Field due to Continuous Charge Distribution JEE Notes | EduRev
dq = ρdV [Charge on infinitely small volume element dV]
Q = ∫dq = ∫ρdV [Total charge on the body]

Steps to calculate Electric Field Intensity due to continuous charge body:
(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
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev

(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.
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev

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

ELECTRIC FIELD CALCULATION DUE TO UNIFORMLY DISTRIBUTED CONTINUOUS CHARGE 
(a) Electric Field on axis of a uniformly charged circular ring:
Electric Field due to Continuous Charge Distribution JEE Notes | EduRevConsider 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
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
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 Electric Field due to Continuous Charge Distribution JEE Notes | EduRev along the x and y axis i.e. Electric Field due to Continuous Charge Distribution JEE Notes | EduRevandElectric Field due to Continuous Charge Distribution JEE Notes | EduRev
By resolvingElectric Field due to Continuous Charge Distribution JEE Notes | EduRevwe get,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Observe and imagine, thatElectric Field due to Continuous Charge Distribution JEE Notes | EduRevwill cancel out if we take each and every element of the ring into consideration.
Therefore net electric field at P,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Now, by geometry,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Thus,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Replacing with the value of λ defined above,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev

(b) Electric field strength at a general point due to a uniformly charged rod:
As shown in figure, if P is any general point in the surrounding of rod, to find the electric field strength at P, again we consider an element on rod of length dx at a distance x from point O as shown in figure.

Electric Field due to Continuous Charge Distribution JEE Notes | EduRev

Now if dE be the electric field at P due to the element, then it can be given as
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Here
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Now we resolve electric field in components. Electric field strength in x-direction due to dq at P is,
dEx = dEsin 𝜃
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Here we have x = r tan 𝜃
and dx = rsec2 𝜃d𝜃
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Net electric field strength due to dq at point P in x-direction is
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
or Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Similarly, the electric field strength at point P due to dq in y-direction is
dEy = dEcos𝜃
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Again we have x = rtan 𝜃
And dx = rsec2 𝜃d𝜃
Thus we have,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Net electric field strength at P due to dq in y-direction is
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
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,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
In perpendicular direction ,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev

We can use this generalized finite relation to calculate the Electric Field due to following systems too:
(i) Infinitely long uniformly charged rod with charge density 𝜆:

Electric Field due to Continuous Charge Distribution JEE Notes | EduRev

For infinite rod, 𝜃1 → 90° and 𝜃2 → 90°
Therefore, for infinitely long uniformly charged rod,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
While,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev

(ii) Electric field due to semi-infinite wire:
For this case,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev

(c) Electric field strength due to a uniformly surface 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.

Electric Field due to Continuous Charge Distribution JEE Notes | EduRev

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
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Here due to the elemental ring electric field strength dE at point P can be given as
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Net electric field at point P due to this disc is given by integrating above expression from O to R as
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Now, using integration by substitution we can solve the above integral as,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
By geometry,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Hence,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Please note that 𝜽 is the angle subtended by the disc at point P which is x distance far from the center.

Case: (i) If x < < R ⇒ cos 𝜽 → 1
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,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
i.e. behaviour of the disc is like infinite sheet.

Case: (ii) If x > > R
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Now, using binomial approximation,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
i.e. behaviour of the disc is like a point charge.

Electric Field Strength due to a uniformly charged Hollow Hemispherical Cup:
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 dq at an angle q from its axis as shown. The surface area of this ring will be
Electric Field due to Continuous Charge Distribution JEE Notes | EduRevdA = 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,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev

Net electric field at centre can be obtained by integrating this expression between limits 0 to 𝜋/2.
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Hence, Electric Field intensity at centre C, due to uniformly charged nonconducting hemispherical shell is,
Electric Field due to Continuous Charge Distribution JEE Notes | EduRev
Above given continuous charged systems are most frequently used ones. It is recommended to remember the procedure and results by heart.

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