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
Let a non-conducting sphere be of radius R with center O. The conducting sphere is uniformly charged. Total charge is q which is uniformly distributed in volume. 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 an extremely small area element dA. The direction of the the 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 non-conducting sphere
2. Electric field on the surface of non-conducting sphere
3. Electric field inside non-conducting sphere
Electric field outside non-conducting sphere (r>R)
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 non-conducting sphere (r = R)
We know that there is r = R on the surface of a conducting sphere.
Hence,
`E_{\text{surface}}` = `\frac{K q}{R^2}` (Maximum Possible Value)
Electric field on the surface of a non-conducting sphere (r < R)
Consider a point P at a distance r fromm its center O in a uniformly charged non-conducting sphere. Total charge is q which is uniformly distributed in total volume. 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 an extremely small area element dA. The direction of the the 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.
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)
Here, `q'` is the charge inside the Gaussian surface
Equate equation 1 and 2
`E \times 4\pi r^2` = `\frac{q'}{\epsilon_0}`
`E_{\text{inside}}` = `\frac{q'}{4 \pi \epsilon_0 r^2}`
`E_{\text{inside}}` = `\frac{1}{4 \pi \epsilon_0}\frac{q'}{r^2}`
`E_{\text{inside}}` = `\frac{K q'}{r^2}` ....Eq. (3)
Calculating `q'`
`q' = \frac{\text{Total Charge}}{\text{Total Volume}}\times\text{Volume of Gaussian surface}`
`q' = \frac{q}{\frac{4}{3}\pi R^3}\times \frac{4}{3}\pi r^3`
`q' = \frac{q}{\frac{4}{3}\pi R^3}\times \frac{4}{3}\pi r^3`
`q' = \frac{q r^3}{R^3}`
By putting the value of `q'` in Equation 3
`E_{\text{inside}}` = `\frac{K q r^3}{r^2 R^3}`
`E_{\text{inside}}` = `\frac{K q r}{ R^3}`
Hence,
`E_{\text{inside}}` `\prop r`
Variation of E with distance r
MCQs
1. According to Gauss's law, the total electric flux passing through a closed surface in an electric field is equal to:
a) The net charge inside the surface
b) The ratio of net charge and permittivity
c) The product of net charge and electric field
d) The product of net charge and surface area
2. What is the relationship between the total electric flux and the net charge inside the closed surface according to Gauss's law?
a) Directly proportional
b) Inversely proportional
c) No relationship
d) Indeterminate
3. The permittivity of vacuum is denoted by:
a) ε
b) `ε_0`
c) μ
d) `μ_0`
4. The equation representing Gauss's law is:
a) φ = Σq
b) φ = (1/`ε_0`)Σq
c) φ = 4πKΣq
d) φ = (4πK/`ε_0`)Σq
5. What is the value of K in Gauss's law equation?
a) `1/(4πε_0)`
b) `4πε_0`
c) `1/(4Ï€K)`
d) `4πKε_0`
6. The electric field due to a non-conducting sphere at a point outside the sphere is given by:
a) `(q)/(4πε_0r^2)`
b) `(Kq)/(4πε_0R^2)`
c) `(Kq)/(4Ï€R^2)`
d) `(Kq)/(4Ï€r^2)`
7. At which point(s) is the electric field maximum for a non-conducting sphere?
a) Outside the sphere
b) On the surface of the sphere
c) Inside the sphere
d) Same at all points
8. The electric field on the surface of a non-conducting sphere is:
a) Zero
b) Maximum
c) Indeterminate
d) Same as the electric field outside the sphere
9. What is the relationship between the electric field inside a non-conducting sphere and the distance from its center?
a) Inversely proportional to the distance
b) Directly proportional to the distance
c) Independent of the distance
d) Inversely proportional to the square of the distance
10. According to Gauss's law, the electric field inside a non-conducting sphere is proportional to:
a) Charge of the sphere
b) Radius of the sphere
c) Both charge and radius of the sphere
d) Neither charge nor radius of the sphere
11. The charge inside a Gaussian surface is given by:
a) `(q r^3)/R^3`
b) `(q r)/R^2`
c) `(q R^2)/r^3`
d) `(q R)/r^2`
12. The electric field outside a non-conducting sphere varies with distance r as:
a) Inversely proportional to r
b) Directly proportional to r
c) Independent of r
d) Inversely proportional to the square of r
Answers:
1. b) The ratio of net charge and permittivity
2. a) Directly proportional
3. b) `ε_0`
4. c) φ = 4πKΣq
5. a) 1/(4π`ε_0`)
6. a) (q)/(`4πε_0r^2`)
7. b) On the surface of the sphere
8. b) Maximum
9. b) Directly proportional to the distance
10. a) Charge of the sphere
11. a) `(q r^3)/R^3`
12. d) Inversely proportional to the square of r
Short Answer type Questions with Answer
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 equation for the total electric flux through a closed surface?
Answer: The equation is: φ = (Σq / ε₀), where φ represents the total flux, Σq is the net charge inside the surface, and ε₀ is the permittivity of vacuum.
3. What is the relationship between ε₀ and K?
Answer: The relationship is: 1 / ε₀ = 4Ï€K.
4. What is the formula for the electric field outside a non-conducting sphere?
Answer: The formula is: `E_{out}` = (Kq / r²), where `E_{out}` is the electric field intensity, q is the total charge, r is the distance from the center of the sphere, and K is a constant.
5. What is the maximum possible value of the electric field on the surface of a non-conducting sphere?
Answer: The maximum value is: `E_{surface}` = (Kq / R²), where `E_{surface}` is the electric field intensity, q is the total charge, and R is the radius of the sphere.
6. What is the formula for the electric field inside a non-conducting sphere?
Answer: The formula is: `E_{\text{inside}}` = (Kqr / R³), where `E_{\text{inside}}` is the electric field intensity, q is the total charge, r is the distance from the center of the sphere, and R is the radius of the sphere.
7. What is the relationship between the electric field inside a non-conducting sphere and the distance from its center?
Answer: The electric field inside a non-conducting sphere varies directly with the distance (r) from its center.
8. How is the total electric flux calculated through a closed surface?
Answer: The total electric flux is calculated by integrating the dot product of the electric field and the area vector over the closed surface.
9. What is the equation for the total electric flux through a closed surface in terms of the electric field and the area?
Answer: The equation is: φ = `\vecE . \vecA`, where φ represents the total flux, E is the electric field, and A is the area vector.
10. What is the charge inside a Gaussian surface used in Gauss's law?
Answer: The charge inside a Gaussian surface is represented by q', and it is the charge enclosed by the surface that is being considered.
Numerical Questions
1. Question: A non-conducting sphere of radius 0.2 m carries a total charge of 5 µC. Calculate the electric field at a distance of 0.5 m from the center of the sphere. (Answer: 5.76 kN/C)
2. Question: A non-conducting sphere has a charge of 10 nC distributed uniformly throughout its volume. Determine the electric field at a distance of 0.1 m from the center of the sphere. (Answer: 35.91 kN/C)
3. Question: A non-conducting sphere of radius 0.1 m carries a charge of 1 µC uniformly distributed on its surface. Calculate the electric field at a point just outside the sphere. (Answer: 36 kN/C)
4. Question: A non-conducting sphere has a charge of -2 µC uniformly distributed on its surface. Find the electric field at a distance of 0.3 m from the center of the sphere. (Answer: 8.72 kN/C)
5. Question: A non-conducting sphere has a charge of 20 µC distributed uniformly throughout its volume. Determine the electric field at a point inside the sphere, at a distance of 0.05 m from the center. (Answer: 1.44 kN/C)
6. Question: A non-conducting sphere of radius 0.3 m carries a charge of 2 µC uniformly distributed on its surface. Calculate the electric field at a distance of 0.1 m from the surface of the sphere. (Answer: 5.38 kN/C)
7. Question: A non-conducting sphere of radius 0.4 m carries a total charge of -8 µC. Determine the electric field at a point inside the sphere, at a distance of 0.3 m from the center. (Answer: 22.47 kN/C)
8. Question: A non-conducting sphere has a charge of 50 nC distributed uniformly throughout its volume. Calculate the electric field at a point just inside the surface of the sphere. (Answer: 4.5 kN/C)
9. Question: A non-conducting sphere of radius 0.5 m carries a charge of -4 µC uniformly distributed on its surface. Find the electric field at a point outside the sphere, at a distance of 1 m from the center. (Answer: 1.44 kN/C)
10. Question: A non-conducting sphere has a charge of 1 µC uniformly distributed throughout its volume. Determine the electric field at a distance of 0.2 m from the center of the sphere. (Answer: 14.4 kN/C)
11. A non-conducting sphere of radius 10 cm has a uniform charge density of 5 μC/m³. Calculate the electric field at a distance of 5 cm from the center of the sphere. (Answer: 8.99 × 10⁴ N/C)
12. A non-conducting sphere of radius R = 10 cm carries a charge density `\sigma = 10^{-9}C/m^3` distributed uniformly throughout its volume. At what distance within the sphere, measured from the center of the sphere, the magnitude of the electric field is E = 1.32 N/m? (Answer: 3.50 cm)
13. Consider a non-conducting sphere with a radius of 8 cm and a charge of 4 μC uniformly distributed throughout its volume. Determine the electric field at a point on the surface of the sphere. (Answer: 2.25 × 10⁴ N/C)
14. A non-conducting sphere has a charge of 10 μC distributed uniformly within it. Calculate the electric field at a distance of 20 cm from the center of the sphere. (Answer: 4.50 × 10⁴ N/C)
15. A non-conducting sphere has a radius of 12 cm and a total charge of 6 μC distributed uniformly within it. Find the electric field at a point 8 cm away from the center of the sphere. (Answer: 6.75 × 10⁴ N/C)
16. Consider a non-conducting sphere with a charge density of 3 μC/m³ and a radius of 6 cm. Determine the electric field at a distance of 4 cm from the center of the sphere. (Answer: 5.99 × 10⁴ N/C)
17. A non-conducting sphere has a charge of 8 μC uniformly distributed within it. Calculate the electric field at a point on the surface of the sphere. (Answer: 2.70 × 10⁴ N/C)
18. A non-conducting sphere has a radius of 15 cm and a total charge of 12 μC distributed uniformly within it. Find the electric field at a distance of 10 cm from the center of the sphere. (Answer: 3.59 × 10⁴ N/C)
19. Consider a non-conducting sphere with a charge density of 2 μC/m³ and a total charge of 5 μC distributed uniformly within it. Determine the electric field at a point 7 cm away from the center of the sphere. (Answer: 6.41 × 10⁴ N/C)
20. A non-conducting sphere has a charge of 15 μC distributed uniformly within it. Calculate the electric field at a point 25 cm away from the center of the sphere. (Answer: 2.16 × 10⁴ N/C)
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