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 Hollow Sphere
Let a conducting sphere be of radius R with center O. The conducting sphere is uniformly charged. Total charge is q which is uniformly distributed on surface. The intensity of the electric field at a point P at a distance r from its center O is to be determined.
To find the electric field intensity, consider a spherical surface passing through a point P, called a Gaussian surface.
Let point P lie on a extremely small area element dA. The direction of extremely small area element dA is in the direction of OP, and the direction of electric field at point P is also in the direction of OP. Thus the extremely small area element and the electric field are in the same direction.
We can find the intensity of the electric field at three different points
1. Electric field outside a spherical shell
2. Electric field on the surface of a spherical shell
3. Electric field inside a spherical shell
Electric field outside a spherical shell (r>R)
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| Electric Field due to Hollow Sphere |
According to definition
`\phi = \oint_s \vec { E}.\vec {dA}` `(dA = dS = \text{area})`
`\phi = \oint_s E dA cos \theta`
`\phi = \oint_s E dA cos 0`
`\phi = \oint_s E dA` `(\because cos 0 = 1)`
`\phi = E \oint_s dA`
`\phi = E A` `(\because \oint dA = A)`
`\phi = E \times 4\pi r^2` ....Eq. (1) `(\because A = 4\pi r^2)`
According to Gauss's law
`\phi = \frac{q}{\epsilon_0}` ....Eq. (2)
Equate equation 1 and 2
`E \times 4\pi r^2` = `\frac{q}{\epsilon_0}`
`E_{\text{out}}` = `\frac{q}{4 \pi \epsilon_0 r^2}`
`E_{\text{out}}` = `\frac{1}{4 \pi \epsilon_0}\frac{q}{r^2}`
`E_{\text{out}}` = `\frac{K q}{r^2}`
In vector form
`\vec E` = `\frac{K q}{r^2}\hat r`
Electric field on the surface of a spherical shell (r = R)
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| Electric field on the surface of a spherical shell |
Hence,
`E_{\text{surface}}` = `\frac{K q}{R^2}` (Maximum Possible Value)
Electric field inside a spherical shell (r < R)
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| Electric field on the surface of a spherical shell |
The value of charge inside a conducting sphere is zero, due to this the electric field intensity inside the conducting sphere is also zero.
`E_{\text{inside}} = 0`
Graph between Electric Field and Distance
1. `E_{\text{out}}` = `\frac{K q}{r^2}`
`E \prop \frac{1}{r^2}`
2. `E_{\text{surface}}` = `\frac{K q}{R^2}`
E = Constant (Maximum possible value)
3. `E_{\text{inside}} = 0`
4. `E_{\text{center}} = 0`
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| Graph between Electric Field and Distance |
Electrostatics Class 12 Chapter 1 Detailed Notes
Electric Field due to Hollow Sphere MCQs
1. According to Gauss's law, the total electric flux passing through a closed surface is equal to:
a) The net charge inside the surface
b) The product of the net charge inside the surface and the permittivity of vacuum
c) The product of the net charge inside the surface and 4Ï€K
d) The product of the net charge inside the surface and 1/ε₀
2. What is the value of ε₀, the permittivity of vacuum?
a) 4Ï€K
b) 1/4Ï€K
c) 1/ε₀
d) 4πε₀
3. The electric field intensity outside a spherical shell is given by:
a) E = Kq/r²
b) E = Kq/R²
c) E = Kq/r
d) E = Kq/R
4. What is the electric field intensity on the surface of a spherical shell?
a) E = Kq/r²
b) E = Kq/R²
c) E = Kq/r
d) E = Kq/R
5. Inside a conducting sphere, the value of the electric field intensity is:
a) E = Kq/r²
b) E = Kq/R²
c) E = 0
d) E = 1/ε₀
6. The graph of electric field intensity versus distance for a point outside a spherical shell follows which relationship?
a) E ∝ 1/r
b) E ∝ r
c) E ∝ 1/r²
d) E ∝ r²
7. Which of the following equations represents Gauss's law?
a) φ = q/ε₀
b) φ = 4πK/q
c) φ = q/4πε₀
d) φ = Kq
8. What is the relationship between the electric flux and the net charge inside a closed surface?
a) φ ∝ q
b) φ ∝ 1/q
c) φ ∝ q²
d) φ ∝ 1/q²
9. When calculating electric flux, what does dA represent?
a) Electric field
b) Electric potential
c) Electric charge
d) Area element
10. The direction of the extremely small area element (dA) and the electric field at a point are:
a) Perpendicular to each other
b) Parallel to each other
c) Opposite to each other
d) Randomly oriented
11. The electric field inside a conducting sphere is zero because:
a) The charges are distributed uniformly
b) The net charge inside the sphere is zero
c) The sphere is made of a non-conductive material
d) The electric field is shielded by the conducting surface
12. What is the maximum possible value of the electric field on the surface of a conducting sphere?
a) E = Kq/r²
b) E = Kq/R²
c) E = Kq/r
d) E = Kq/R
13. What is the electric field intensity at a point P on the
surface of a conducting sphere?
a) E = Kq/r²
b) E = Kq/R²
c) E = Kq/r
d) E = Kq/R
14. At the center of a conducting sphere, the electric field intensity is:
a) E = Kq/r²
b) E = Kq/R²
c) E = 0
d) E = 1/ε₀
Answers:
1. d) The product of the net charge inside the surface and 1/ε₀
2. b) 1/4Ï€K
3. a) E = Kq/r²
4. b) E = Kq/R²
5. c) E = 0
6. c) E ∝ 1/r²
7. a) φ = q/ε₀
8. a) φ ∝ q
9. d) Area element
10. b) Parallel to each other
11. b) The net charge inside the sphere is zero
12. b) E = Kq/R²
13. b) E = Kq/R²
14. c) E = 0
Short Answer type Question
1. What is Gauss's law?
Answer: Gauss's law states that the total electric flux passing through a closed surface in an electric field is equal to the product of the net charge inside the surface and the inverse of the permittivity of vacuum.
2. What is the mathematical expression for Gauss's law?
Answer: The mathematical expression for Gauss's law is: `\phi = \frac{\Sigma q}{\epsilon_0}`, where `\phi` represents the electric flux, `\Sigma q` is the net charge inside the closed surface, and `\epsilon_0` is the permittivity of vacuum.
3. What is the relationship between `\epsilon_0` and `K`?
Answer: The relationship between `\epsilon_0` and `K` is: `\frac{1}{\epsilon_0} = 4 \pi K`.
4. How can the total electric flux be calculated using Gauss's law?
Answer: The total electric flux can be calculated using the formula: `\phi = 4 \pi K \times \Sigma q`.
5. What is the direction of the electric field and the area element on Gaussian Surface?
Answer: The electric field and the area element have the same direction on Gaussian Surface.
6. What are the three different points for which the intensity of the electric field is calculated?
Answer: The three different points for calculating the intensity of the electric field are: outside a spherical shell, on the surface of a spherical shell, and inside a spherical shell.
7. What is the expression for the electric field outside a spherical shell?
Answer: The expression for the electric field outside a spherical shell is: `E_{\text{out}} = \frac{K q}{r^2}`, where `r` is the distance from the center of the shell.
8. What is the expression for the electric field on the surface of a spherical shell?
Answer: The expression for the electric field on the surface of a spherical shell is: `E_{\text{surface}} = \frac{K q}{R^2}`, where `R` is the radius of the shell.
9. What is the electric field inside a spherical shell?
Answer: The electric field inside a spherical shell is zero.
10. What is the relationship between electric field and distance outside a spherical shell?
Answer: The relationship between electric field and distance outside a spherical shell is: `E \propto \frac{1}{r^2}`.
11. What is the maximum possible value of the electric field on the surface of a spherical shell?
Answer: The maximum possible value of the electric field on the surface of a spherical shell is given by `E_{\text{surface}} = \frac{K q}{R^2}`.
12. What is the electric field inside a conducting sphere?
Answer: The electric field inside a conducting sphere is zero.
13. What is the graph between electric field and distance outside a spherical shell?
Answer: The graph between electric field and distance outside a spherical shell shows an inverse square relationship: `E \propto \frac{1}{r^2}`.
14. What is the electric field at the center of a conducting sphere?
Answer: The electric field at the center of a conducting sphere is zero.
15. What is the value of charge inside a conducting sphere?
Answer: The value of charge inside a conducting sphere is zero.
Numerical Question
1. Question: A conducting sphere of radius 5 cm carries a charge of 8 μC. What is the electric field at a point located 10 cm away from the center of the sphere? (Answer: 2.304 kN/C)
2. Question: A conducting sphere with a charge of -4 nC is surrounded by a concentric Gaussian surface of radius 12 cm. Calculate the electric field at this surface. (Answer: E ≈ -7.195 x 10⁸ N/C)
3. Question: A conducting sphere has a charge of 10 μC. Determine the electric field at a point 15 cm away from the center of the sphere. (Answer: 4.44 kN/C)
4. Question: The electric field at a distance of 6 cm from the center of a conducting sphere is found to be 12 kN/C. What is the charge on the sphere? (Answer: 3.38 μC)
5. Question: A conducting sphere with a charge of -2 μC is enclosed by a Gaussian surface of radius 8 cm. Calculate the electric field at this surface. (Answer: -45 kN/C)
6. Question: The electric field at a distance of 4 cm from the center of a conducting sphere is 9 kN/C. Determine the charge on the sphere. (Answer: 4.08 μC)
7. Question: A conducting sphere of radius 6 cm has a charge of 12 μC. Calculate the electric field at a point located 12 cm away from the center of the sphere. (Answer: 0.6 kN/C)
8. Question: A conducting sphere with a charge of -6 nC is surrounded by a Gaussian surface of radius 10 cm. Find the electric field at this surface. (Answer: -21.6 kN/C)
9. Question: The electric field at a distance of 8 cm from the center of a conducting sphere is 18 kN/C. Determine the charge on the sphere. (Answer: 13.5 μC)
10. Question: A conducting sphere of radius 4 cm carries a charge of 5 μC. What is the electric field at a point located 6 cm away from the center of the sphere? (Answer: 1.25 kN/C)
PHYSICS NOTES
- Electric Charge, Basic Properties of Electric Charge, Conductors and Insulators, and Methods of Charging
- Continuous Charge Distribution
- Coulomb's Law, Electroscope, Properties of Charge, Quantization of Charge.
- Coulomb's Law in Vector Form
- Principle of Superposition of Charges
- Electric Field and Field Lines, Types of Electric Field, Electric Field Due to a Point Charge, Electric Filed Due to a System of Charges, Electric Field lines and their Properties
- Gauss's Law of Electrostatic, Definition of Gauss's Law, Formula, Electric Charge and its Four Properties, Applications of Garss's Law, Gauss's Law, and Important Points
- Prove of Gauss Theorem in Electrostatics
- Electric Field due to Conducting Sphere
- Electric Field due to Conducting Hollow Sphere
- Electric Field due to Non-Conducting Sphere
- Electric Field due to Infinitely Long Charged Wire
- Electric Field due to Infinite Conducting Sheet of Charge..
- Electric Field due to Non-Conducting Plane Sheet
- Electric Flux
- Definitions of Electric Dipole, Electric Field due to Electric Dipole on Axial Line..
- Electric Field Due to Electric Dipole on Equatorial Line
- Torque on a Dipole in Uniform Electric Field
- Work Done in Rotating a Dipole..
- Dielectric Constant
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