Electric Field due to Uniformly Charged Infinite Plane non-conducting Sheet

Introduction to Gauss's Law

    In this post, you will learn about  Gauss's Law, its derivation, electric flux, and key applications in electrostatics for Class 12 Physics. This topic is highly important for board exams, JEE, and NEET, as it simplifies electric field calculations using symmetry and the Gaussian surface. Here, we will calculate the electric field for a uniformly charged infinite non-conducting sheet.

What is Gauss's Law?

Statement of Gauss's law

    Electric flux through a closed surface equals the net enclosed charge divided by `\epsilon_0.`

`\phi = \frac{\Sigma q_{\text{enclosed}}}{\epsilon_0}`

Gauss's law does not calculate the electric field directly; it simplifies the calculation using symmetry.

Key Core Principles

Fundamental Relation

    Connects the charge distribution to the electric fields by relating the electric flux to enclosed charge.

        `\phi = \frac{q}{\epsilon_0}`

    Flux depends on the component of the electric field normal to the surface.

Shape Independence

    Total flux through a closed surface depends only on the enclosed charge, not the shape or size of the surface.

Simplification of Electric Fields

    Helps calculate electric fields easily in symmetric cases like spherical, cylindrical, and planar systems.

Physical Interpretation

Positive charges act as sources (field lines outward) and negative charges as sinks (field lines inward), which total flux depending only on enclosed charge.

Electric Flux

Definition

Electric flux through a surface is defined as the dot product of the electric field and the area vector.

`\phi = \vec E.\vec A = EAcos\theta`

Where,

`\phi = `Electric flux

`\vec E = `Electric field

`\vec A = `Area vector (perpendicular to surface)

`\theta = `angle between `\vec E` and `\vec A`

Mathematical Form of Gauss's Law

Standard Equation

Thus, the total flux

`\phi = \frac{\Sigma q}{\epsilon_0}`        ......eq.(1)

Here 

`\Sigma q ` is the algebraic sum of charges that exist inside the surface.

`\epsilon_0` is the permittivity of vacuum

Relation With Coulomb's Constant

We know that

`\frac{1}{4 \pi \epsilon_0} = K`

`\frac{1}{ \epsilon_0} = 4 \pi K`

Then from Equation 1

`\phi = \frac{q}{\epsilon_0}`

Alternate Form

`\phi = 4 \pi K \times \Sigma q`

Electric Field due to a Non-Conducting Sheet

    Consider a uniformly charged non-conducting sheet with a surface charge density of `\sigma`.

• The electric field at a point near the sheet is independent of distance.

• The direction of the electric field is perpendicular to the non-conductor sheet.

Uniformly Charged Infinite Plane Sheet Diagram

Assumptions for Infinite Plane Sheet

• The sheet is considered infinite in extent, so edge effects are neglected.

• The electric field is uniform and perpendicular to the sheet.

• The electric field has the same magnitude on both sides of the sheet.

Due to symmetry, the electric field is normal to the surface and equal on both faces of the Gaussian cylinder.

Consider a cylindrical Gaussian surface intersecting the sheet.

1.    Circular surface `S_1.`

2.    Circular surface `S_2.`

3.    Curved Surface `S_3.`

Flux Calculation

    Total outgoing flux from all three surfaces (or we can say from the Gaussian surface).

`\phi_E = \oint_s \vec E.d\vec A = \frac{q'}{\epsilon_0}`

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

`\int_{s_1} E  dA  cos 0^\circ+ \int_{s_2} E  dA  cos 0^\circ + \int_{s_3} E  dA  cos 90^\circ = \frac{\sigma A}{\epsilon_0}`        `( q' = \sigma A)`

`\int_{s_1} E  dA  (1)+ \int_{s_1} E  dA  (1) + \int_{s_3} E  dA  (0) = \frac{\sigma A}{\epsilon_0}`        `(\because s_1 = s_2)`

`\int_{s_1} E  dA (1)+ \int_{s_1} E dA (1) + 0 = \frac{\sigma A}{\epsilon_0}`

`\int_{s_1} E dA + \int_{s_2} E dA  + 0 = \frac{\sigma A}{\epsilon_0}`

`\int_{s_1} 2 E  dA = \frac{\sigma A}{\epsilon_0}`

`2 E \int_{s_1} dA = \frac{\sigma A}{\epsilon_0}`

`2 E  A  = \frac{\sigma A}{\epsilon_0}`        `(\because \int_{s_1} dA = A)`

`2 E  A   = \frac{\sigma A }{\epsilon_0}`

`2 E   = \frac{\sigma }{\epsilon_0}`

`E  = \frac{\sigma }{2 \epsilon_0}`

Here,

`\sigma = \text{Charge density}`

`\sigma = \frac{\text{Charge}}{\text{Area}}`

`q' = \text{Charge inside Gaussian Surface}`

`q' = \sigma A`

In vector form

`\vec E  = \frac{\sigma }{2 \epsilon_0} \hat n`

    Where `\hat n` is the unit vector perpendicular to the sheet.

Definitions

        `\sigma = \frac {\text{Charge}}{\text{Area}}`

        `q'= \sigma A`

Physical Interpretation

• Flux depends only on enclosed charge.

• Outside charges do not contribute.

• The shape of the surface does not matter.

Conclusion -

• Formuls: Electric field `\E = \frac{\sigma}{2\epsilon_0}.`

• Produces a uniform electric field.

• The electric field is independent of distance.

E vs r graph showing a constant electric field (horizontal line) for an infinite plane sheet.

Frequently Asked Questions (FAQ)

Conceptual FAQ

Q.    Why is Gauss's law more useful in symmetry cases?

Q.    Why does the field not depend on distance for an infinite sheet?

Q.    What does Gauss's Law actually tell us?

Q.    Why do we use a Gaussian surface?

Q.    Is Gauss's Law valid for any shape of surface?

Q.    Does a charge outside the surface affect flux?

Exam FAQ

Q.    What is surface charge density?

Q.    What is a closed surface?

Q.    What is the definition of a Gaussian surface?

Q.    Why do we use imaginary surfaces?

Applications of Gauss's Law

    Gauss's Law is mainly used to find the electric field in symmetric cases.

• To find the electric field of symmetrical charge distributions.
• Electric field due to an infinite plane sheet.
• Electric field due to an infinite line charge.
• Electric field due to a spherical shell.
• Electric field due to a uniformly charged solid sphere.
• 
  

Numerical Questions

1. A uniformly charged infinite plane sheet has a surface charge density of 2 C/m². Calculate the electric field produced by this sheet. (Answer: 1.13 × 10¹¹ N/C)

2. A charge of 5 μC is distributed over an area of 10 m² on an infinite plane sheet. Determine the electric field strength at a point near the sheet. (Answer: 2.83 × 10⁻⁷ N/C)

3. The surface charge density of a uniformly charged plane sheet is 4 × 10⁻⁴ C/m². What is the electric field it generates? (Answer: 2.26 × 107 N/C)

4. An infinite plane sheet carries a total charge of 8 μC distributed uniformly. If the area of the sheet is 4 m², find the electric field produced. (Answer: 2 × 10⁻⁶ N/C)

5. The electric field due to a uniformly charged infinite plane sheet is measured to be 3.5 × 10⁻⁹ N/C. If the surface charge density is 7 μC/m², calculate the permittivity of air. (Answer: 8.33 × 10⁻¹2 C²/(Nm²))

6. A plane sheet has a surface charge density of 1.5 × 10⁻⁶ C/m². If the electric field generated by this sheet is 9.6 × 10⁻¹⁰ N/C, determine the area of the sheet. (Answer: 6.25 m²)

7. The electric field produced by a uniformly charged infinite plane sheet is 2.5 × 10⁻⁸ N/C. If the charge distributed on the sheet is 4 μC, what is the surface charge density? (Answer: 1.6 × 10⁻⁶ C/m²)

8. An infinite plane sheet has a surface charge density of 3.2 × 10⁻⁵ C/m². If the electric field near the sheet is 1.6 × 10⁻⁹ N/C, determine the area of the sheet. (Answer: 2 × 10⁴ m²)

9. The electric field due to a uniformly charged infinite plane sheet is 6.4 × 10⁻¹² N/C. If the area of the sheet is 5 m², calculate the total charge distributed on the sheet. (Answer: 3.2 × 10⁻¹¹ C)

10. A plane sheet with a surface charge density of 5 × 10⁻⁸ C/m² generates an electric field of 1.2 × 10⁻⁹ N/C. Determine the permittivity of air. (Answer: 2.4 × 10⁻¹² C²/(Nm²))

Call to Action (CTA)

    If you found this post helpful, don't stop here - take your preparation to the next level with structured resources and guided learning.

• Free Important Question PDF Download

• Join Physics365 WhatsApp-Get daily questions & updates. Join Now

• Watch Full Video- Physics 365

👉Don't just read, start practicing and boost your score today!

Read More