By Mohammed Saeed

7 November 2019

In the previous tutorial we learned how to find the Electric Field due to a charge distribution.

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Gauss's Law is another way to find the Electric Field

Gauss's Law can be used to solve complex electrostatic problems

Gauss’s Law can be used to simplify evaluation of electric field in a simple way.

Gauss's law states that the total electric flux ψ through any closed surface is equal to the total charge enclosed by that surface.ψ = 𝑄𝑒𝑛𝑐= 𝑫. 𝑑𝑠 = ρ𝑣 dv.

Differential normal area is 𝑑𝑠 = 𝑑𝑠 𝑛 where dS is the area of the surface element and 𝑛 is a unit vector normal to the surface dS

After Finding the electric flux density 𝑫, the electric field can be calculated from 𝑫 = 𝑬𝜀𝑜

We apply Gauss’s Law in the following way: Choose a Gaussian surface, such that evaluation of electric field becomes easy

Make use of symmetry to make problems easier

Remember, it is not necessary that Gaussian surface to match with real surface, it can be inside or outside the Gaussian surface. Also, it can be an imaginary geometric surface. So, the only requirement imposed on a Gaussian surface is that it be closed.

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In this tutorial we will cover these applications:

Electric Field due to Infinite Line

Electric Field due thin Spherical Shell

Outside the spherical shell

Inside the spherical shell

Electric Field due to Infinite Plate Sheet

Electric Field due to Spherical

Outside the spherical

Inside the spherical

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Find the electric field due to an infinite line charge distribution 𝜌𝑙

Solution

Step 1.

- We select the surface to be an infinite cylinder centered by the line charge

- Because it is an infinite line, the field crossing the upper and lower surface = 0, and the electric charge density is constant (𝐷 = 𝐷𝜌 𝑎 𝜌). In another word, D has same value for any point on side surface of the cylinder.

Step 2.

- Apply Gauss’s law on a length L of the cylinder

- Note: 𝜌 is the radius of the cylinder 𝜌𝑙 is charge distribution

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𝜌𝑙𝐿 2𝜋𝜌𝐿

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Three Spherical shells r=1m, r=2m, and r=3m have charge distributions ρ

𝑠1, ρ

𝑠2,and ρ

𝑠3are 2, -4, 5 µc/𝑚2 respectively

a) Calculate the flux pass through r=0.5m, r=1.5m, and r=2.5mb) Find 𝑫 at r=0.5m, r=2.5m, and r=3.5m

Solution

At r=1m ρ𝑠1= 2µc/𝑚2, At r=2m ρ𝑠2=-4µc/𝑚2, At r=3m ρ𝑠3=5 µc/𝑚2

a) Electric Flux ψ1. At r=0.5m

ψ = 𝑄𝑒𝑛𝑐=0 (because no charge inside the Gauss’s surface)

2. At r=1.5m

ψ = 𝑄𝑒𝑛𝑐= 𝑄1 = ρ𝑠1ds = ρ𝑠1

𝐴𝑠𝑝ℎ𝑒𝑟𝑒 = ρ𝑠1

4π𝑟2 = 2*10−6 (4π12) = 8π µc

3. At r=2.5m

ψ = 𝑄𝑒𝑛𝑐= ρ𝑠1ds + ρ𝑠2ds = ρ𝑠1

4π𝑟2 + ρ𝑠2

4π𝑟2 = 2*10−6 (4π12) - 4*10−6 (4π22) = -56π µc

b) Electric Flux Density 𝑫1. At r=0.5m

𝑄𝑒𝑛𝑐 = 𝑫. 𝑑𝑠 = 0 So 𝑫=0

2. At r=2.5m

𝑄𝑒𝑛𝑐 = 𝑫. 𝑑𝑠 = 𝐷𝑟 𝑑𝑠 = 𝐷𝑟 (4π𝑟2)= 𝐷𝑟 (4π2.52) = -56π µc

So,𝐷𝑟= - 2.24µc/𝑚2

r=1m

r=2m

r=3m

Why r=2m and not 2.5m?

Why r=2.5m and not 2m?Why negative?

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r=1m

r=2m

r=3m

Three Spherical shells r=1m, r=2m, and r=3m have charge distributions ρ

𝑠1, ρ

𝑠2,and ρ

𝑠3are 2, -4, 5 µc/𝑚2 respectively

a) Calculate the flux pass through r=0.5m, r=1.5m, and r=2.5mb) Find 𝑫 at r=0.5m, r=2.5m, and r=3.5m

Solution

b) Electric Flux Density 𝑫

3. At r=3.5m

𝑄𝑒𝑛𝑐 = 𝑄1 + 𝑄2 + 𝑄3 = ρ𝑠1ds + ρ𝑠2ds + ρ𝑠3ds = -56π*10−6 + 5*10−6(4π𝑟2) = -56π*10−6 + 5*10−6(4π32) = 124π µc

Then we can find 𝑫 from Gauss’s law

𝑄𝑒𝑛𝑐 = 𝑫. 𝑑𝑠 = 𝐷𝑟 𝑑𝑠 = 𝐷𝑟 (4π𝑟2)= 𝐷𝑟 (4π3.52) = 124π µc

So,

𝐷𝑟= 2.531 µc/𝑚2

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Solution

a) Electric Flux ψ1. At r=2m

ψ = 𝑄𝑒𝑛𝑐= ρ𝑣dv = 𝑟=0

2 θ=0

π ϕ=0

2π(10

𝑟2 ∗ 10−3).(𝑟2sinθ dr dθ dϕ) = 2π

25C

2. At r=6m

ψ = 𝑄𝑒𝑛𝑐= ρ𝑣dv = 𝑟=0

4 θ=0

π ϕ=0

2π(10

𝑟2 ∗ 10−3).(𝑟2sinθ dr dθ dϕ) = 0.16π C

b) Electric Flux Density 𝑫1. At r=1m

𝑄𝑒𝑛𝑐= ρ𝑣dv = 𝑟=0

1 θ=0

π ϕ=0

2π(10

𝑟2 ∗ 10−3).(𝑟2sinθ dr dθ dϕ) = π

25C

Then, from Gauss’s law

𝑄𝑒𝑛𝑐 = 𝑫. 𝑑𝑠 = 𝐷𝑟 𝑑𝑠 = 𝐷𝑟 (4π𝑟2)= 𝐷𝑟 (4π12) = π

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𝐷𝑟= 10 mc/𝑚2 8

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2. At r=5m

ψ = 𝑄𝑒𝑛𝑐= ρ𝑣dv = 𝑟=0

4 θ=0

π ϕ=0

2π(10

𝑟2 ∗ 10−3).(𝑟2sinθ dr dθ dϕ) = 0.16π C

Then, from Gauss’s law

𝑄𝑒𝑛𝑐 = 𝑫. 𝑑𝑠 = 𝐷𝑟 𝑑𝑠 = 𝐷𝑟 (4π𝑟2)= 𝐷𝑟 (4π52) = 0.16π C

𝐷𝑟= 1.6 mc/𝑚2

# Note

𝑄𝑒𝑛𝑐 is the same for r=5m and 6m, because in both cases the same amount of charge is enclosed

However,

the strength of 𝑫 on a certain point on the sphere depends on the distance r

In another word, 𝑫 at r=5m deferent than 𝑫 at r=6m

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The problem of a coaxial cable is almost identical with that of the line charge.

The charge density on the inner conductor (ρ𝑠)

The charge density on the outer conductor (ρ𝑠)

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A 50-cm length of coaxial cable has an inner radius of 1 mm and an outer radius of 4 mm. The space between conductors is assumed to be filled with air. The total charge on the inner conductor is 30 nC. ①Find the charge density on each conductor

② the expressions for E and D fields.

Solution ①

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Now to find D and E, we will assume the Gaussian surface as a cylinder of length L and radios ρ (Note ρ is a variable). There are three possibilities:

1. When a < ρ < b. Then from Gauss’s law we can find D and E

Note that: from the last equations we can conclude that when ρ=a , then 𝐷𝜌 = ρ𝑠,𝑖𝑛𝑛𝑒𝑟 𝑐𝑦𝑙

2. When ρ > b

Then the total charge enclosed will be Zero, because one conducting carry a positive charge and the other negative charge with same magnitude. As a result D=E=0

3. When ρ < a

Then the total charge enclosed will be zero, because there is no charge. As a result D=E=011

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Find the electric field strength 𝐸 because of infinite sheet with surface charge Density 𝜌𝑠 and placed on the plane z = 0

Solution

Step 1

- We set the Gaussian surface as a rectangular box with the sheet at the middle (Also it can be a cylinder)

- Since the sheet extends to infinity then 𝑫 on the sides isalways zero

- 𝑫 exists on the upper and lower sides 𝑫= 𝐷𝑧 𝑎𝑧 and 𝑫= −𝐷𝑧 𝑎𝑧 respectively.

Step 2

- Apply Gauss’s law

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- Then we can write 𝑫 and 𝑬 on the top and bottom sides as

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Note

This problem is not like the second question in this tutorial

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