Electric Field Due to Linear Charge Distribution Class 12 | Gauss Law Derivation, Formula & Notes

    Physics365.in brings you a complete guide to the electric field due to a line charge for Class 12, including a step-by-step derivation of Gauss's law, the linear charge density formula, and detailed electrostatics notes.

Concept of Linear Charge Distribution

 A linear charge distribution means charge is spread along a one-dimensional object.

Examples include a wire.

It is described using linear charge density (\lambda).`

This concept helps us calculate electric fields using Coulomb's law.

Examples of Linear Charge Distribution

Charged Thin Rod

A uniformly charged straight rod where the charge is spread evenly along its length.

Long Straight Wire

An infinitely long wire carrying a uniform charge is commonly used in Gauss's law problems.

Circular Ring of Charge

A ring where charge is distributed along its circumference (treated as a linear distribution).

Linear Charge Density `(\lambda)`

Linear charge density `(\lambda)` is the charge per unit length, given by `\lambda = q/L`, measured in `C/m`, used for charge on thin wires or rods.

Formula

`\lambda = q/L`

Where:

`\lambda =` linear charge density

`q =` total charge 

`L =` length of the rod

    Now that we understand linear charge density, let's move on to deriving the electric field.

Electric Field Derivation

Step 1: Choose Gaussian Surface

    Let there be an infinitely long line charge extending from -∞ to +∞. We need to find the electric field intensity at a point P, located at a perpendicular distance r from the line charge.

    To determine the electric field at point P, we consider an imaginary cylindrical surface coaxial with the line charge passing through point P. This surface is known as a Gaussian surface.

Diagram of the electric field due to a line charge

Gauss’s Law: Electric Field of an Infinite Line Charge

    The Gaussian surface is divided into an upper circular surface `S_1`, a lower circular surface `S_2`, and a curved surface `S_3`.

The Gaussian surface may be divided into three parts -

1.    `S_1` (top cap): flux = 0 (electric field ⊥ area vector)

2.    `S_2` (bottom cap): flux = 0

3.    `S_3` (Curved Surface): contributes fully

Reason: The electric field is parallel to the area only on the curved surface, so it contributes fully to the electric flux.

Application of Gauss's Law

    The total electric flux through the Gaussian surface is given by:

`\phi_E = \oint_S \vec E.d \vec A = \frac{\Sigma q}{\epsilon_0}`

Step 2: Splitting the Flux Over Surfaces

`\int_{s_1} \vec E.d\vec A + \int_{s_2}\vec E.d\vecA + \int_{s_3}\vec E.d\vecA = \frac{\Sigma q}{\epsilon_0}`

`\int_{s_1} E  dA  cos 90^\circ + \int_{s_2} E  dA  cos 90^\circ + \int_{s_3} E  dA  cos 0^\circ = \frac{\lambda l}{\epsilon_0}`        `(\Sigma q = \lambda l)`

`\int_{s_1} E  dA  (0)+ \int_{s_2} E  dA  (0) + \int_{s_3} E  dA  (1) = \frac{\lambda l}{\epsilon_0}`

`0+ 0 + \int_{s_3}E dA = \frac{\lambda l}{\epsilon_0}`

`0 + 0 +E  \int_{s_3}  dA  = \frac{\lambda l}{\epsilon_0}`

`E \int_{s_3}  dA  = \frac{\lambda l}{\epsilon_0}`

`E  2 \pi r l  = \frac{\lambda l}{\epsilon_0}`        `(\because \int_{S_3} = 2\pirl)`

`E  2 \pi r   = \frac{\lambda }{\epsilon_0}`

Step 3: Final Formula

`E  = \frac{\lambda }{2 \pi \epsilon_0 r}`

This represents the total electric flux through the curved surface.

`E  = \frac{2 \lambda }{4 \pi \epsilon_0 r}`

Scalar Form of Electric Field

`E  = \frac{2 K \lambda }{r}`        `(\because K = \frac{1}{4 \pi \epsilon_0})`

Vector Form of Electric Field

`\vec E = \frac {2 K \lambda}{r} \hat r`

`\vec E = \frac{1}{4 \pi \epsilon_0}\frac {2 \lambda}{r} \hat r`

    Where `\hat r` is a unit vector in the direction of OP and perpendicular to the line charge.

Relation between E and r

`E \propto \frac{1}{r}`

Electric field (E) varies inversely with distance (r) from a line charge (E ∝ 1/r)

Quick Revision

• `\lambda = q/L`

• `E = \frac{\lambda}{2\pi\epsilon_0 r}`

• `E\propto 1/r`

MCQs on Linear Charge Distribution

    Now, let's test your understanding with some important MCQs.

1. Gauss's law relates the total electric flux passing through a closed surface to which of the following quantities?

   A) Electric potential

   B) Electric field strength

   C) Net charge inside the surface

   D) Permittivity of vacuum

2. Why is the electric flux through the top and bottom surfaces of a cylindrical Gaussian surface zero?

   A) Electric field is zero

   B) Area is zero

  C) Electric field is perpendicular to the area vector

  D) The electric field is parallel to the area vector

3. What is the angle between the electric field vector and the area vector for the lower circular surface of the Gaussian surface?

   A) 0 degrees

   B) 45 degrees

   C) 90 degrees

   D) 180 degrees

4. What is the angle between the electric field vector and the area vector for the curved surface of the Gaussian surface?

   A) 0 degrees

   B) 45 degrees

   C) 90 degrees

   D) 180 degrees

5. The unit vector `\hat{r}` in the expression for the electric field represents:

    A) The direction of electric potential

    B) The direction of net charge

    C) The direction of the electric field

    D) The direction perpendicular to the line charge

6. What is the relationship between the electric field strength `\vec{E}` and the distance r from the line charge?

    A) `\vec{E} \propto r`

    B) `\vec{E} \propto \frac{1}{r}`

    C) `\vec{E} \propto r^2`

    D) `\vec{E} \propto \frac{1}{r^2}`

7. The value of the constant K in the expression for the electric field is equal to:

    A) `\frac{1}{4\pi \epsilon_0}`

    B) `\frac{1}{2\pi \epsilon_0}`

    C) `4\pi \epsilon_0`

    D) `2\pi \epsilon_0`

8. Gauss's law is applicable to which type of surfaces?

    A) Open surfaces

    B) Closed surfaces

    C) Planar surfaces

    D) Spherical surfaces

Answers:

1. C) Net charge inside the surface

2. C) Electric field is perpendicular to the area vector

3. C) 90 degrees

4. C) 90 degrees

5. D) The direction perpendicular to the line charge

6. B) `\vec{E} \propto \frac{1}{r}`

7. A) `\frac{1}{4\pi \epsilon_0}`

8. B) Closed surfaces

Frequently Asked Questions (FAQ)

1. How is the electric field (E) related to the distance (r) from the linear charge?

Answer: The electric field (E) is inversely proportional to the distance (r) from the linear charge. `E \propto \frac{1}{r}`.

Numerical Questions with Answers

1. A closed surface contains zero net charge. What is the total electric flux passing through the surface? (Answer: 0 Nm²/C)

2. A Gaussian surface around an infinite line charge with a linear charge density of 0 μC/m. If the distance from the line charge to the Gaussian surface is 2 m, what is the intensity of the electric field at that point? (Answer: 0 N/C)

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