Gauss's law
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`.
Intensity of Electric Field due to Infinite Linear Charge
Let there be a line charge of infinite length extending from + infinity to – infinity. We have to find the intensity of electric field at a point P at a distance r perpendicular to it. To find the intensity of the electric field at point P, imagine a cylindrical surface passing through the point P. It is called a Gaussian surface.
 |
| Electric Field due to Linear Charge Distribution |
The Gaussian surface is divided into an upper circular surface `S_1`, a lower circular surface `S_2`, and a curved surface `S_3`.
This Gaussian surface may be divided into three parts -
1. Upper circular surface `S_1`
2. Lower 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{\Sigma q}{\epsilon_0}`
`\oint_s \vec E.d\vec A = \frac{\Sigma q}{\epsilon_0}`
`\int_{s_1} \vec E.d\vec A + \int_{s_2}\vecE.d\vecA + \int_{s_3}\vecE.d\vecA = \frac{\Sigma q}{\epsilon_0}`
`\int_{s_1} E dA cos 0^\circ+ \int_{s_2} E dA cos 90^\circ + \int_{s_3} E dA cos 90^\circ = \frac{\lamda l}{\epsilon_0}` `(\Sigma q = \lamda l)`
`\int_{s_1} E dA (1)+ \int_{s_2} E dA (0) + \int_{s_3} E dA (0) = \frac{\lamda l}{\epsilon_0}`
`\int_{s_1} E dA (1)+ 0 + 0 = \frac{\lamda l}{\epsilon_0}`
`E \int_{s_1} dA + 0 + 0 = \frac{\lamda l}{\epsilon_0}`
`E \int_{s_1} dA = \frac{\lamda l}{\epsilon_0}`
`E 2 \pi r l = \frac{\lamda l}{\epsilon_0}` `(\because \int_{s_1} = 2 \pi r l)`
`E 2 \pi r = \frac{\lamda }{\epsilon_0}`
`E = \frac{\lamda }{2 \pi \epsilon_0 r}`
`E = \frac{2 \lamda }{4 \pi \epsilon_0 r}`
`E = \frac{2 K \lamda }{r}` `(\because K = \frac{1}{4 \pi \epsilon_0})`
In vector form
`\vec E = \frac {2 K \lamda}{r} \hat r`
`\vec E = \frac{1}{4 \pi \epsilon_0}\frac {2 \lamda}{r} \hat r`
Where `\hat r` is unit vector in the direction of OP and perpendicular to line charge.
Relation between E and r
`E \prop \frac{1}{r}`
 |
| Electric Field due to Linear Charge Distribution |
Electrostatics Class 12 Chapter 1 Detailed Notes
MCQs on Electric field due to Linear Charge Distribution
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. What is the relationship between the permittivity of vacuum `\epsilon_0` and the constant K?
A) `\epsilon_0 = K`
B) `K = 4\pi \epsilon_0`
C) `\epsilon_0 = \frac{1}{4\pi K}`
D) `K = \frac{1}{\epsilon_0}`
3. The Gaussian surface used to calculate the electric field due to an infinite linear charge has three parts. Which of the following is NOT one of those parts?
A) Upper circular surface
B) Lower circular surface
C) Curved surface
D) Spherical surface
4. The total outgoing flux from the Gaussian surface is equal to which of the following quantities?
A) Electric field strength
B) Electric potential
C) Net charge inside the surface
D) Permittivity of vacuum
5. What is the angle between the electric field vector and the area vector for the upper circular surface of the Gaussian surface?
A) 0 degrees
B) 45 degrees
C) 90 degrees
D) 180 degrees
6. 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
7. 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
8. The expression for the electric field `\vec{E}` due to an infinite linear charge and charge density `\lambda` at a distance r is given by:
A) `\vec{E} = \frac{\lambda}{4\pi \epsilon_0 r}\hat r`
B) `\vec{E} = \frac{\lambda}{2\pi \epsilon_0 r}\hat r`
C) `\vec{E} = \frac{2\lambda}{4\pi \epsilon_0 r}\hat r`
D) `\vec{E} = \frac{2\lambda}{r}\hat r`
9. 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
10. 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}`
11. 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`
12. Gauss's law is applicable to which type of surfaces?
A) Open surfaces
B) Closed surfaces
C) Planar surfaces
D) Spherical surfaces
13. The total electric flux passing through a closed surface is equal to the product of the net charge inside the volume of the closed surface and:
A) `4\pi K`
B) `\epsilon_0`
C) Electric field strength
D) Permeability of vacuum
Answers:
1. C) Net charge inside the surface
2. B) `K = 4\pi \epsilon_0`
3. D) Spherical surface
4. C) Net charge inside the surface
5. C) 90 degrees
6. C) 90 degrees
7. A) 0 degrees
8. C) `\vec{E} = \frac{ 2\lambda}{4\pi \epsilon_0 r}`
9. D) The direction perpendicular to the line charge
10. B) `\vec{E} \propto \frac{1}{r}`
11. A) `\frac{1}{4\pi \epsilon_0}`
12. B) Closed surfaces
13. A) `4\pi K`
No comments:
Post a Comment