Electric Field due to Infinite Conducting Sheet of Charge

 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 Sphere

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

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.

    The Gaussian surface is divided into an two circular surface `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} 0  dA  cos\theta + \int_{s_3} E  dA  cos 90^\circ = \frac{\sigma A}{\epsilon_0}`        `( q' = \sigma A)`

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

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

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

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

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

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

`E  = \frac{\sigma }{\epsilon_0}`

Here,

`\sigma = \text{Charge density}`

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

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

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

`q' = \sigma A`

In vector form

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

    Where `\hat n` is unit vector.

Electrostatics Class 12 Chapter 1 Detailed Notes

Questions and Answers

Q.1         What is Gauss’s Law in terms of total electric flux through a closed surface in a vacuum of air?

Ans :      According to Gauss’s Law the total electric flux through a closed surface is equal to the product of the net charge inside the surface and ` \frac{1}{\epsilon_0}`.

Where,

 `\phi = \frac{q}{\epsilon_0}`

q = Charge inside the Gaussian surface

 ` \epsilon_0` = permittivity of vacuum

 

Q.2         What does `\Sigma q`   represent in Gauss’s Law?

Ans :      `\Sigma q` is the algebraic sum of the charges inside the closed surface.

 

Q.3         What is ` \epsilon_0` in Gauss’s Law?

Ans :       is the permittivity of a vacuum.  is a constant that represents the ability of a vacuum to permit electric field lines.

 

Q.4         How is the constant K related to ` \epsilon_0` ?

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

 

Q.5         What is the electric field due to a uniformly charged conducting sheet with a surface charge density  ?

Ans :     `E\ =\ \frac{\sigma}{\epsilon_0}`

 

Q.6         What is the vector form of the electric field due to the conducting surface (sheet)?

Ans :       `\vec{E}\ =\ \frac{\sigma}{\epsilon_0}\ \hat{n}`        Where, ` \hat{n}` is the unit vector perpendicular to the surface.

 

Q.7         What is the direction of the electric field due to a uniformly charged conducting sheet?

Ans :      The direction of the electric field due to a uniformly charged conducting sheet is perpendicular to the surface of the sheet and pointing away for positive charge.

 

Q.8         How does the electric field due to a conducting sheet vary with distance?

Ans :       `\vec{E}\ =\ \frac{\sigma}{\epsilon_0}\ \hat{n}`     Where, ` \hat{n}` is the unit vector perpendicular to the surface.

Independent of distance from the sheet. Remains constant.