Electric Field due to Uniformly Charged Infinite Plane non conducting Sheet

Gauss's law

    The total electric flux passing through a closed surface kept in an electric field in vacuum or air is equal to the product of the net charge `\Sigma q` inside the volume of the closed surface and `\frac{1}{\epsilon_0}`

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 a permittivity of vacuum

We know that

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

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

then from Equation 1

`\phi = \frac{1}{\epsilon_0}\times \Sigma q`

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

    The total electric flux passing through a closed surface kept in an electric field in a vacuum or air is equal to the product of the net charge `\Sigma q` inside the volume of the closed surface and `\phi = 4 \pi K`.

Electric Field due to non-Conducting Sheet

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

The electric field at point P due to this sheet at a distance r is to be determined.

The direction of electric field intensity at point P is in the perpendicular direction to the conductor sheet.

Electric Field due to Uniformly Charged Infinite Plane Sheet

    The Gaussian surface is divided into two circular surfaces `S_1`, a and `S_2`, and a curved surface `S_3`.

This Gaussian surface may be divided into three parts -

1.    Circular surface `S_1`

2.    Circular surface `S_2`

3.    Curved Surface `S_3`

    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}`

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

`\int_{s_1} \vec E.d\vec A + \int_{s_2}\vecE.d\vecA + \int_{s_3}\vecE.d\vecA = \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_1} 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 `\hatn` is the unit vector perpendicular to the sheet.

Conclusion -

`\star`    It produces a uniform electric field.

`\star`    Electric field due to infinite uniformly charged sheet does not depend upon distance.

Electric Field due to Uniformly Charged Infinite Plane Non Conducting Sheet

Electrostatics Class 12 Chapter 1 Detailed Notes

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 × 10⁻¹¹ 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 × 10⁻⁸ 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 permitivity of air. (Answer: 8.33 × 10⁻¹³ 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 permitivity of air. (Answer: 2.4 × 10⁻¹² C²/(Nm²))