An electron initially 3.00 m from a nonconducting infinite sheet of uniformly distributed charge is fired toward the sheet. The electron has an initial speed of 390 m/s and travels along a line perpendicular to the sheet. When the electron has traveled 2.00 m , its velocity is instantaneously zero, and it then reverses its direction.

Required:
a. What is the surface charge density on the sheet?
b. Given the same initial velocity, from what distance should the electron be fired if it is to just reach the sheet?

Answer:

60.8 cm²

Explanation:

The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².

σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²

Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.

Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface

So, q = ε₀Ф = σA'

ε₀Ф = σA'

making A' the area of the Gaussian surface the subject of the formula, we have

A' = ε₀Ф/σ

A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²

A' = 81.4568/1.34 × 10⁻⁴ m²

A' = 60.79 × 10⁻⁴ m²

A' ≅ 60.8 cm²

The flux through the Gaussian surface is 9.20 N⋅m^2/C then the surface area of the Gaussian Sheet is 60.76 square cm.

A certain amount of electrons in excess or defect is called a charge. Charge density is the amount of charge distributed over per unit of volume.

Given that, for a thin sheet of insulating material, the charge Q is 87.6 pC and surface area A is 65.2 square cm. Then the charge density for a thin sheet is given below.

[tex]\sigma = \dfrac {Q}{A}[/tex]

[tex]\sigma = \dfrac {87.6\times 10^{-12}}{65;.2\times 10^{-4}}[/tex]

[tex]\sigma = 1.34\times 10^{-8} \;\rm C/m^2[/tex]

Thus the charge density for a thin sheet of insulating material is [tex]1.34\times 10^{-8} \;\rm C/m^2[/tex].

Now, the flux through the Gaussian surface is 9.20 N⋅m^2/C. The charge over the Gaussian Surface is given as below.

[tex]Q' = \sigma A'[/tex]

Where Q' is the charge at the Gaussian Surface, A' is the surface area of the Gaussian surface and [tex]\sigma[/tex] is the charge density.

For the closed Gaussian Surface, Flux is given below.

[tex]\phi = \dfrac {Q'}{\epsilon_\circ}[/tex]

Hence the charge can be written as,

[tex]Q' = \phi\epsilon_\circ[/tex]

So the charge can be given as below.

[tex]Q' = \phi\epsilon_\circ = \sigma A'[/tex]

Then the surface area of the Gaussian surface is given below.

[tex]A' = \dfrac {\phi\epsilon_\circ}{\sigma}[/tex]

Substituting the values in the above equation,

[tex]A' = \dfrac {9.20 \times 8.85\times 10^{-12}}{1.38\times 10^{-8}}[/tex]

[tex]A' =0.006076\;\rm m^2[/tex]

[tex]A' = 60.76 \;\rm cm^2[/tex]

Hence we can conclude that the area of the Gaussian Surface is 60.76 square cm.

To know more about the charge and charge density, follow the link given below.

https://brainly.com/question/8532098.

Answer:

θ₀ = 84.78° (OR) 5.22°

Explanation:

This situation can be treated as projectile motion. The parameters of this projectile motion are:

R = Range of Projectile = 150 m

V₀ = Launch Speed of Projectile = 90 m/s

g = 9.8 m/s²

θ₀ = Launch angle (OR) Angle of Elevation = ?

The formula for range of a projectile is given as:

R = V₀² Sin 2θ₀/g

Sin 2θ₀ = Rg/V₀²

Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²

2θ₀ = Sin⁻¹ (0.18)

θ₀ = 10.45°/2

θ₀ = 5.22°

Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:

θ₀ = 90° - 5.22°

θ₀ = 84.78°

Answer:

(a) Distance between deer and car = 5 m

(b) Vmax = 21.92 m/s

Explanation:

a.

First we calculate distance covered during response time:

s₁ = vt   --------- equation 1

where,

s₁ = distance covered during response time = ?

v = speed of car = 20 m/s

t = response time = 0.5 s

Therefore,

s₁ = (20 m/s)(0.5 s)

s₁ = 10 m

Now, we calculate the distance covered by the car during deceleration. Using 3rd equation of motion:

2as₂ = Vf² - Vi²

s₂ = (Vf² - Vi²)/2a ------ eqation 2

where,

a = deceleration = - 10 m/s²

s₂ = Distance covered during deceleration = ?

Vf = Final Velocity = 0 m/s (since car finally stops)

Vi = Initial Velocity = 20 m/s

Therefore,

s₂ = [(0 m/s)² - (20 m/s)²]/2(-10 m/s²)

s₂ = (400 m²/s²)/(20 m/s²)

s₂ = 20 m

thus, the total distance covered by the car before coming to rest is given as:

s = s₁ + s₂

s = 10 m + 20 m

s = 30 m

Now, the distance between deer and car, when it comes to rest, can be calculated as:

Distance between deer and car = 35 m - s = 35 m - 30 m

Distance between deer and car = 5 m

b.

Since, the distance covered by the car in total must be equal to 35 m at maximum. Therefore,

s₁ + s₂ = 35 m

using equation 1 and equation 2 from previous part:

Vi t + (Vf² - Vi²)/2a = 35 m

Vi(0.5 s) + [(0 m/s)² - Vi²]/2(-10 m/s²) = 35 m

0.5 Vi + 0.05 Vi² = 35

0.05 Vi² + 0.5 Vi - 35 = 0

solving this quadratic equation, we get:

Vi = - 31.92 m/s  (OR)  Vi = 21.92 m/s

For maximum velocity:

Vmax = 21.92 m/s

Answer:

I believe it is they will weigh the same

Explanation:

Center of gravity is the axis on which the mass rotates evenly if I remember correctly from AP Physics

The head side is heavier than the handle side. - this will be discovered.

Theoretically, the body's center of gravity is where all of the weight is believed to be concentrated. Knowing the centre of gravity is crucial because it may be used to forecast how a moving object will behave when subjected to the effects of gravity. In designing immobile constructions like buildings and bridges, it is also helpful.

We know that center of gravity is  close to some particular point refers the mass of the point is greater then others. It is given that: The center of gravity of an ax is on the centerline of the handle, close to the head.

So, we can conclude that the head side of the ax is heavier than the handle side of it.

Learn more about center of gravity here:

https://brainly.com/question/17409320

#SPJ5

Answer:

Explanation:

Given data

length of cube= 0.2015 m

density = 19300 kg/m^3.

But the volume of cube is given as [tex]l*l*l= l^3[/tex]

[tex]volume -of- cube= 0.2015*0.2015*0.2015= 0.00818 m^3[/tex]

The density is expressed as = mass/volume

[tex]mass=19300*0.00818= 157.87m^3[/tex]

Complete Question

The uniform purely axial magnetic induction required by the experiment in a volume large enough to accommodate the Lorentz Tube is produced by the Helmholtz Coils. What is the magnetic induction due to a coil current 1.5 Ampere? Convert the result in the still popular non-SI unit Gauss (1 Tesla = 10^4 Gauss).

B = N*mue*I/(2*r)

# of loops = 140

radius of the coil = 0.14m

Answer:

 The magnetic induction is [tex]B = 2.639 \ Gauss[/tex]

Explanation:

From the question we are told that

     The coil current is  [tex]I = 1.5 \ A[/tex]

     The number of loops is  [tex]N = 140[/tex]

The magnetic field due to the current is mathematically represented as

           [tex]B = \mu_o * N * I[/tex]

[tex]\mu_o[/tex] is the permeability of free space with value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

substituting value

           [tex]B = 4\pi * 10^{-7} * 140 * 1.5[/tex]

           [tex]B = 2.639*19^{-4} \ T[/tex]

From question

        (1 Tesla = [tex]10^4 \ Gauss[/tex]).

=>      [tex]B = 2.693 *10^{-4} *10^4 = 2.63 \ Gauss[/tex]

=>      [tex]B = 2.639 \ Gauss[/tex]

Answer:

The apparent weight of the pilot is 3311 N

Explanation:

Given;

radius of the vertical circle, r = 600 m

speed of the plane, v = 150 m/s

mass of the pilot, m = 70 kg

Weight of the pilot due to his circular motion;

[tex]W= F_v\\\\F_v = \frac{mv^2}{r} \\\\F_v = \frac{70*150^2}{600} \\\\F_v = 2625 \ N[/tex]

Real weight of the pilot;

[tex]W_R = mg\\\\W_R = 70 *9.8\\\\W_R = 686 \ N[/tex]

Apparent weight - Real weight of pilot = weight due to centripetal force

[tex]F_N - mg = \frac{mv^2}{r} \\\\F_N = \frac{mv^2}{r} + mg\\\\F_N = 2625 \ N + 686 \ N\\\\F_N = 3311\ N[/tex]

Therefore, the apparent weight of the pilot is 3311 N

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  [tex]f_e = 0.027 \ m[/tex]

Explanation:

From the question we are told that

    The focal length is  [tex]f_o = 5.5 \ mm = -0.0055 \ m[/tex]

This negative sign shows the the microscope is diverging light

     The  angular magnification is [tex]m = 300[/tex]

     The  distance between the objective and the eyepieces lenses is  [tex]Z = 19 \ cm = 0.19 \ m[/tex]

Generally the magnification is mathematically represented as

        [tex]m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ][/tex]

Where [tex]f_e[/tex] is the eyepiece focal length of the microscope

  Now  making [tex]f_e[/tex] the subject  of the formula

         [tex]f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }[/tex]

substituting values

        [tex]f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }[/tex]

         [tex]f_e = 0.027 \ m[/tex]

Answer:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = [tex]\frac{1}{2}ke^{2}[/tex]

==> [tex]\frac{1}{2}* 550* 0.02^{2}[/tex] = 0.11 J

Answer:

  K_{f} / K₀ =1.12

Explanation:

This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.

Initial moment. With arms outstretched

         L₀ = I₀ w₀

the wo value is 5.0 rad / s

final moment. After he shrugs his arms

         [tex]L_{f}[/tex] = I_{f}  w_{f}

indicate that the moment of inertia decreases by 11%

        I_{f} = I₀ - 0.11 I₀ = 0.89 I₀

        L_{f} = L₀

        I_{f} w_{f}  = I₀ w₀

        w_{f} = I₀ /I_{f}    w₀

let's calculate

        w_{f} = I₀ / 0.89 I₀   5.0

        w_{f} = 5.62 rad / s

Having these values ​​we can calculate the change in kinetic energy

         [tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)

         K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²

         K_{f} / K₀ =1.12

Answer:

A. Her total angular momentum has decreased

Explanation:

Total angular momentum is the product of her moment of inertia and angular velocity. In this scenario it doesn’t decrease but rather remains constant as the movement of the arms doesn’t have any effect on the total angular momentum.

The movement of the arm under certain conditions however has varying effects and changes on parameters such as the moment of inertia and the angular speed.

Answer:

16 years.

Explanation:

Using Kepler's third Law.

P2=D^3

P=√d^3

Where P is the orbital period and d is the distance from the sun.

From the question the semi major axis of the asteroid is 4 AU= distance. The distance is always express in astronomical units.

P=?

P= √4^3

P= √256

P= 16 years.

Orbital period is 16 years.

Answer:

The wavelength will be 33.9 cm

Explanation:

Given;

frequency of the wave, F = 1200 Hz

Tension on the wire, T = 800 N

wavelength, λ = 39.1 cm

[tex]F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}[/tex]

Where;

F is the frequency of the wave

T is tension on the string

μ is mass per unit length of the string

λ is wavelength

[tex]\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\[/tex]

when the tension is decreased to 600 N, that is T₂ = 600 N

[tex]T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm[/tex]

Therefore, the wavelength will be 33.9 cm

Answer:

The distance between the two charges is =4.4mm

Answer:

the correct answer is d Beats

Explanation:

when two sound waves interfere time has different frequencies, the result is the sum of the waves is

       y = 2A cos 2π (f₁-f₂)/2    cos 2π (f₁ + f₂)/2

where in this expression the first part represents the envelope and the second part represents the pulse or beatings of the wave.

When examining the correct answer is d Beats

Answer:

c.they have opposite charges.

Explanation:

because the protons have a positive charge and the neutrons have no charge.

Answer:

The astronaut and the satellite are 53.718 m apart.

Explanation:

Given;

mass of spacewalking astronaut, = 88 kg

mass of satellite, = 645 kg

force exerts by the satellite, F = 110N

time for this action, t = 0.45 s

Determine the acceleration of the satellite after the push

F = ma

a = F / m

a = 110 / 645

a = 0.171 m/s²

Determine the final velocity of the satellite;

v = u + at

where;

u is the initial velocity of the satellite = 0

v = 0 + 0.171 x 0.45

v = 0.077 m/s

Determine the displacement of the satellite after 1.4 m

d₁ = vt

d₁ = 0.077 x (1.4 x 60)

d₁ = 6.468 m

According to Newton's third law of motion, action and reaction are equal and opposite;

Determine the backward acceleration of the astronaut after the push;

F = ma

a = F / m

a = 110 / 88

a = 1.25 m/s²

Determine the final velocity of the astronaut

v = u + at

The initial velocity of the astronaut = 0

v = 1.25 x 0.45

v = 0.5625 m/s

Determine the displacement of the astronaut after 1.4 min

d₂ = vt

d₂ = 0.5625 x (1.4 x 60)

d₂ = 47.25 m

Finally, determine the total separation between the astronaut and the satellite;

total separation = d₁ + d₂

total separation = 6.468 m + 47.25 m

total separation = 53.718 m

Therefore, the astronaut and the satellite are 53.718 m apart.

Answer:

Ф = 28.9°

Explanation:

given:

radius (r) = 117m

velocity (v) = 25.1 m/s

required: angle Ф

Ф = inv tan (v² / (r * g))      we know that g = 9.8

Ф = inv tan (25.1² / (117 * 9.8))

Ф = 28.9°

Answer:

6 atm

Explanation:

PV = PV

(4 atm) (3 L) = P (2 L)

P = 6 atm

Given info

d = 0.000250 meters = distance between slits

L = 302 cm = 0.302 meters = distance from slits to screen

[tex]\theta_8 = 1.12^{\circ}[/tex] = angle to 8th max (note how m = 8 since we're comparing this to the form [tex]\theta_m[/tex])

[tex]x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}[/tex] (n = 5 as we're dealing with the 5th minimum )

---------------

Method 1

[tex]d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}[/tex]

Make sure your calculator is in degree mode.

-----------------

Method 2

[tex]\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\[/tex]

-----------------

Method 3

[tex]\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\[/tex]

There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.

Answer:

Explanation:

Heat absorbed by oil

= mass x specific heat x rise in temperature

= 30 x 10⁻³ x 950 x 2.2 x 10³ x ( 50-10 )

= 25.08 x 10⁵ J  

Heat absorbed by air

= 50 x 1.2 x 1.0054 x 10³ x ( 20-10 )

= 6.03 x 10⁵ J

Total heat absorbed = 31.11 x 10⁵ J

If time required = t

heat lost from room

= .35 x 10³ t

Total heat generated in time t

= 1.8 x 10³ t

Heat generated = heat used

1.8 x 10³ t =  .35 x 10³ t  + 31.11 x 10⁵

1.45 x 10³ t = 31.11 x 10⁵

t = 31.11 x 10⁵ / 1.45 x 10³

t = 2145.5 s

Answer:

The speed will be "3.4×10⁴ m/s²".

Explanation:

The given values are:

Angular speed,

w = 7200 rpm

i.e.,

  = [tex]7200 \times \frac{2 \pi}{60}[/tex]

  = [tex]753.6 \ rad/s[/tex]

Speed from the center,

r = 6.0 cm

As we know,

⇒  Linear speed, [tex]v=wr[/tex]

On putting the estimated values, we get

                               [tex]=753.6\times 0.06[/tex]

                               [tex]=45.216 \ m[/tex]

Now,

Acceleration on disk will be:

⇒  [tex]a=\frac{v^2}{r}[/tex]

       [tex]=34074 \ m/s^2[/tex]

       [tex]=3.4\times 10^4 \ m/s^2[/tex]

Answer:

superscript

Explanation:

When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.

An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.

Answer:

superscript

Explanation:

Answer:

The maximum speed of the mass is 4.437 m/s.

Explanation:

Given;

length of pendulum, L = 1.62 m

attached mass, m = 117 g

angle of inclination, θ = 38°

This mass was raised to a height of

h = 1.62 - cos38° = 1.0043 m

Apply the principle of conservation of mechanical energy

PE = KE

mgh = ¹/₂mv²

v  = √(2gh)

v = √(2 * 9.8 * 1.0043)

v = 4.437 m/s.

Therefore, the maximum speed of the mass is 4.437 m/s.

Answer:

he correct answers are a, b

Explanation:

In the two-slit interference phenomenon, the expression for interference is

          d sin θ= m λ                       constructive interference

          d sin θ = (m + ½) λ             destructive interference

in general this phenomenon occurs for small angles, for which we can write

           tanθ = y / L

           tan te = sin tea / cos tea = sin tea

           sin θ = y / La

un

derestimate the first two equations.

Let's do the calculation for constructive interference

         d y / L = m λ

the distance between maximum clos is and

         y = (me / d) λ

this is the position of each maximum, the distance between two consecutive maximums

         y₂-y₁ = (L   2/d) λ - (L 1 / d) λ₁          y₂ -y₁ = L / d λ

examining this equation if the wavelength decreases the value of y also decreases

the same calculation for destructive interference

         d y / L = (m + ½) κ

         y = [(m + ½) L / d] λ

again when it decreases the decrease the distance

the correct answers are a, b

Answer:

the two beams must travel paths that differ by one-half of a wavelength.

Answer:

95.5 m

Explanation:

The displacement is the position of the ending point relative to the starting point.

In this case, the magnitude of the displacement is the diameter of the circular track.

d = 300 m / π

d ≈ 95.5 m

Answer:

The initial velocity of spaceship P was u₁ = 5.06 m/s

Explanation:

In an elastic collision between two bodies the expression for the final velocity of the second body is given as follows:

[tex]V_{2} = \frac{(m_{2}-m_{1}) }{(m_{1}+m_{2})}u_{2} + \frac{2m_{1} }{(m_{1}+m_{2})}u_{1}[/tex]

Here, subscript 1 is used for spaceship P and subscript 2 is used for spaceship T. In this equation:

V₂ = Final Speed of Spaceship T = 8.1 m/s

m₁ = mass of spaceship P = 4 M

m₂ = mass of spaceship T = M

u₁ = Initial Speed of Spaceship P = ?

u₂ = Initial Speed of Spaceship T = 0 m/s

Using these values in the given equation, we get:

[tex]8.1 m/s = \frac{M-4M }{4M+M}(0 m/s) + \frac{2(4M) }{4M+M}u_{1}[/tex]

8.1 m/s = (8 M/5 M)u₁

u₁ = (5/8)(8.1 m/s)

u₁ = 5.06 m/s

Answer:

3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²

- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²

Explanation:

Complete Question

3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?

- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.

The image of this setup attached to this question as obtained from online is attached to this solution.

Solution

3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.

Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²

The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²

- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.

I₂ = I₁ cos² θ

where

I₂ = intensity of light that passes through the second polarizer = ?

I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²

θ = angle between the major axis of the first and second polarizer = 30°

I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²

In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through

I₃ = I₂ cos² θ

I₃ = intensity of light that passes through the third/additional polarizer = ?

I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²

θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)

I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²

Hope this Helps!!!

Answer:

the answers are provided in the attachments below

Explanation:

Gauss law state that the net electric field coming out of a closed surface is directly proportional to the charge enclosed inside the closed surface

Applying Gauss law to the long solid cylinder

A) E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]

B) E = 2K λ / r

C) Answers from parts a and b are the same

D) attached below

Applying Gauss's law which states that the net electric field in an enclosed surface is directly ∝ to the charge found in the enclosed surface.

A ) The expression for the electric field inside the volume at a distance r

Gauss law :  E. A = [tex]\frac{q}{e_{0} }[/tex]  ----- ( 1 )

where : A = surface area = 2πrL ,  q = p(πr²L)

back to equation ( 1 )

E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]

B) Electric field at point Outside the volume in terms of charge per unit length  λ

Given that:  linear charge density = area * volume charge density

                                            λ    =  πR²P

from Gauss's law : E ( 2πrL) = [tex]\frac{q}{e_{0} }[/tex]

∴ E = [tex]\frac{\pi R^{2}P }{2e_{0}r\pi }[/tex]    ----- ( 2 )

where : πR²P = λ

Back to equation ( 2 )

E = λ  / 2e₀π*r              where : k = 1 / 4πe₀

∴ The electric field ( E ) at point outside the volume in terms of charge per unit Length λ

E = 2K λ / r

C) Comparing answers A and B

Answers to part A and B are similar

Hence we can conclude that Applying Gauss law to the long solid cylinder

E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex], E = 2K λ / r also Answers from parts a and b are the same.

Learn more about Gauss's Law : https://brainly.com/question/15175106