Ask question

Ask question

All categories

Thepotemich [5.8K]

3 years ago

15

It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an

altitude of 300 m the field has magnitude 60.0 N/C; at an altitude of 230 m, the magnitude is 100 N/C. Find the amount of charge contained in a cube 70 m on edge, with horizontal faces at altitudes of 230 and 300 m. Neglect the curvature of Earth.

1 answer:

Lemur [1.5K]3 years ago

7 0

Answer:

q=1.7346×10⁻⁶C

Explanation:

Since the electric field is perpendicular to the bottom and top of the cube,the total flux is equals the flux over the top of surface plus the flex over the lower surface

Ф(total)=Ф₃₀₀+Ф₂₃₀

But the flux is given by Ф=E.A=EACos(θ) where θ is the angle between Area vector and electric field

So

Ф(total)=E₃₀₀A Cos(180)+E₂₃₀ACos(0)

Ф(total)=A(E₃₀₀ - E₂₃₀)

The total flux is given by Gauss Law as:

Ф(total)=q/ε₀

q=ε₀Ф(total)

q=ε₀(A(E₃₀₀ - E₂₃₀))

Substitute the given values

q=(8.85×10⁻¹²){(70²)(100 - 60)}

q=1.7346×10⁻⁶C

You might be interested in

A 150 kg car hits a wall with 45 N of force, what was its acceleration?

alisha [4.7K]

Answer:

a = 0.3 m/s²

Explanation:

Given: 45 N, 150 kg

To find: a

Formula: $a = \frac{F}{m}$

Solution: To find a, divide the force by the weight

A = F ÷ m

= 45 ÷ 150

= 0.3 m/s²

Newtons are derived units, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.

4 0

3 years ago

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

Strike441 [17]

Answer:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$

Explanation:

The electric field equation of a electromagnetic wave is given by:

$E=E_{max}(kx-\omega t)$ (1)

E(max) is the maximun value of E, it means the amplitude of the wave.
k is the wave number
ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

$k=\frac{2\pi}{\lambda}$

$k=8.98*10^{6} [rad/m]$

And the relation between λ and f is:

$c=\lambda f$

$f=\frac{c}{\lambda}$

$f=\frac{3*10^{8}}{700*10^{-9}}$

$f=4.28*10^{14}$

The angular frequency equation is:

$\omega=2\pi f$

$\omega=2\pi*4.28*10^{14}$

$\omega=2.69*10^{15} [rad/s]$

Therefore, the E equation, suing (1), will be:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$ (2)

For the magnetic field we have the next equation:

$B=B_{max}(kx-\omega t)$ (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

$E_{max}=cB_{max}$

$B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}$

$B_{max}=1.17*10^{-8}T$

Putting this in (3), finally we will have:

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$ (4)

I hope it helps you!

8 0

3 years ago

Which of these are part of our solar system? select all that apply

Len [333]

A)planets
b)the sun
c)moons
e)comets
f)asteroids

6 0

3 years ago

Read 2 more answers

1. State the law of conservation of energy and what it means for you as a human considering how energy works.

Arturiano [62]

Answer: 1. The law of consevation of energy sates that energy can neither be created nor destroyed. It can only be transformed or transfered from one form to another. The law of conservation of energy is found everywhere for example, Water falls from the sky, converting potential energy to kinetic energy.

2. Different forms of energy are related because energy cannot be created or destroyed. they can all be transformed into from one form to another.

Explanation:

5 0

3 years ago

1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.

NikAS [45]

Answer:

1.

109.6 cm , - 1.74 , real

2.

1.5

Explanation:

1.

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{40} = \frac{1}{63} + \frac{1}{d}$

d = 109.6 cm

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{-109.6}{63}$

m = - 1.74

The image is real

2

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}$

a = 10

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{- (- (a +5))}{a}$

$m = \frac{(5 + 10)}{10}$

m = 1.5

6 0

3 years ago