Electric Field Due To A Line Charge

• Electric Field Due To A Line Charge

Consider a thin infinitely long straight line charge of linear charge density λ.
Let P be the point at a distance ‘a’ from the line. To find electric field at point P, draw a cylindrical surface of radius ‘a’ and length l.
If E is the magnitude of electric field at point P, then electric flux through the Gaussian surface is given by,

Φ = E × Area of the curved surface of a cylinder of radius r and length l

Because electric lines of force are parallel to end faces (circular caps) of the cylinder, there is no component of field along the normal to the end faces.

Φ = E × 2πal … (i)
According to Gauss theorem, we have

From equations (i) and (ii), we obtain

• Electric Field Due To An Infinite Plane Sheet Of Charge
  

Consider an infinite thin plane sheet of positive charge having a uniform surface charge density σon both sides of the sheet. Let P be the point at a distance ‘a’ from the sheet at which electric field is required. Draw a Gaussian cylinder of area of cross-section A through point P.
The electric flux crossing through the Gaussian surface is given by,

Φ = E × Area of the circular caps of the cylinder

Since electric lines of force are parallel to the curved surface of the cylinder, the flux due to electric field of the plane sheet of charge passes only through the two circular caps of the cylinder.

Φ = E × 2A … (i)
According to Gauss theorem, we have

Here, the charge enclosed by the Gaussian surface,
q = σA

From equations (i) and (ii), we obtain

• Electric Field Due To A Uniformly Charged Thin Spherical Shell
  

• When point P lies outside the spherical shell
  

Suppose that we have to calculate electric field at the point P at a distance r (r > R) from its centre. Draw the Gaussian surface through point P so as to enclose the charged spherical shell. The Gaussian surface is a spherical shell of radius r and centre O.
Let be the electric field at point P. Then, the electric flux through area element is given by,

Since is also along normal to the surface,
dΦ = E ds
∴ Total electric flux through the Gaussian surface is given by,

Now,

Since the charge enclosed by the Gaussian surface is q, according to Gauss theorem,

From equations (i) and (ii), we obtain

• When point P lies inside the spherical shell
  

In such a case, the Gaussian surface encloses no charge.
According to Gauss law,
E × 4πr² = 0
i.e., = E = 0 (r < R)

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BASIC PROPERTIES OF ELECTRIC CHARGES

Basic Properties of Electric Charges

• Additive nature of charges − The total electric charge on an object is equal to the algebraic sum of all the electric charges distributed on the different parts of the object. If q₁, q₂, q₃, … are electric charges present on different parts of an object, then total electric charge on the object, q = q₁ + q₂ + q₃ + …
  

• Charge is conserved − When an isolated system consists of many charged bodies within it, due to interaction among these bodies, charges may get redistributed. However, it is found that the total charge of the isolated system is always conserved.
  
• Quantization of charge − All observable charges are always some integral multiple of elementary charge, e (= ± 1.6 × 10⁻¹⁹ C). This is known as quantization of charge.
  

Coulomb’s Law

• Two point charges attract or repel each other with a force which is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
  

Where, [In SI, when the two charges are located in vacuum]
− Absolute permittivity of free space = 8.854 × 10⁻¹² C² N⁻¹ m⁻²

We can write equation (i) as

• The force between two charges q₁ and q₂ located at a distance r in a medium may be expressed as
  

Where − Absolute permittivity of the medium

The ratio is denoted by ε_r, which is called relative permittivity of the medium with respect to vacuum. It is also denoted by k, called dielectric constant of the medium.

ε = kε₀

Coulomb’s Law in Vector Form

Consider two like charges q₁ and q₂ present at points A and B in vacuum at a distance r apart.
According to Coulomb’s law, the magnitude of force on charge q₁ due to q₂ (or on charge q₂ due to q₁) is given by,

Let
− Unit vector pointing from charge q₁ to q₂
− Unit vector pointing from charge q₂ to q₁
[is along the direction of unit vector ] …(ii)
[is along the direction of unit vector] …(iii)

∴Equation (ii) becomes

On comparing equation (iii) with equation (iv), we obtain

Forces between Multiple Charges
Principle of superposition − Force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges.

Consider that n point charges q₁, q₂, q₃, … q_n are distributed in space in a discrete manner. The charges are interacting with each other. Let the charges q₂, q₃, … q_n exert forces on charge q₁. Then, according to principle of superposition, the total force on charge q₁ is given by,

If the distance between the charges q₁ and q₂ is denoted as r₁₂; and is unit vector from charge q₂ to q₁, then

Similarly, the force on charge q₁ due to other charges is given by,


Substituting these in equation (i),

Electric Field − It is the space around a charge, in which any other charge experiences an electrostatic force.
Electric Field Intensity − The electric field intensity at a point due to a source charge is defined as the force experienced per unit positive test charge placed at that point without disturbing the source charge.

Where,
→ Electric field intensity
Force experienced by the test charge q₀
Its SI unit is NC⁻¹.
Electric Field Due To a Point Charge

We have to find electric field at point P due to point charge +q placed at the origin such that
To find the same, place a vanishingly small positive test charge q₀ at point P.
According to Coulomb’s law, force on the test charge q₀ due to charge q is

If is the electric field at point P, then

The magnitude of the electric field at point P is given by,

Representation of Electric Field

Electric Field Due To a System of Charges

Consider that n point charges q₁, q₂, q₃, … q_n exert forces on the test charge placed at origin O.
Let be force due to i^th charge q_i on q₀. Then,

Where, r_i is the distance of the test charge q₀ from q_i
The electric field at the observation point P is given by,

If is the electric field at point P due to the system of charges, then by principal of superposition of electric fields,

Using equation (i), we obtain

Electric Field Lines
An electric line of force is the path along which a unit positive charge would move, if it is free to do so.

• Properties of Electric Lines of Force
  

• These start from the positive charge and end at the negative charge.
  
• They always originate or terminate at right angles to the surface of the charge.
  
• They can never intersect each other because it will mean that at that particular point, electric field has two directions. It is not possible.
  
• They do not pass through a conductor.
  
• They contract longitudinally.
  
• They exert a lateral pressure on each other.
  

• Representation of Electric Field Lines
  

• Field lines in case of isolated point charges
  

• Field lines in case of a system of two charges
  

  Continuous Charge Distribution

• Linear charge density − When charge is distributed along a line, the charge distribution is called linear.
  

Where,
λ → Linear charge density
q → Charge distributed along a line
L → Length of the rod

• Surface charge density
  

Where,
σ → Surface charge density
q → Charge distributed on area A

• Volume charge density
  The electric flux, through a surface, held inside an electric field represents the total number of electric lines of force crossing the surface in a direction normal to the surface.
  Electric flux is a scalar quantity and is denoted by Φ.
  
  
  SI unit − Nm² C⁻¹
  Gauss Theorem
  It states that the total electric flux through a closed surface enclosing a charge is equal to times the magnitude of the charge enclosed.
  
  However,
  ∴Gauss theorem may be expressed as
  
  Proof
  
  Consider that a point electric charge q is situated at the centre of a sphere of radius ‘a’.
  According to Coulomb’s law,
  
  Where, is unit vector along the line OP
  The electric flux through area element is given by,
  
  
  Therefore, electric flux through the closed surface of the sphere,
  
  It proves the Gauss theorem in electrostatics.
  

Where,
δ → Volume charge density
V → Volume of the conductor
q → Charge on conductor