Ask question

Ask question

All categories

miss Akunina [59]

3 years ago

7

A cubical shell with edges of length a is positioned so that two adjacent sides of one face are coincident with the +x and +y ax

es of a rectangular coordinate system and the corner formed by these two sides is at the origin. An electric field of magnitude E=bx2 directed along the +x axis exists in this region.
1.How much charge is contained in the volume of the shell?
2.Express your answer in terms of the variables a, b, and the permittivity constant ϵ0.

1 answer:

fomenos3 years ago

3 0

Answer: Q = ba⁴ * ε₀

Explanation:

Remember, Gauss's Law states that:

"The net electric flux through any hypothetical closed surface is equal to 1/ε₀ times the net electric charge within that closed surface" i.e

flux Φ = Q / ε₀ where

ε₀ = 8.85e-12 C²/N·m²

Also, the flux, Φ = EAcosθ, where

E = magnitude of the electric field in V/m

A = area of the surface in m²

θ = angle between the electric field lines and the normal (perpendicular) to S.

The field is directed along the x-axis, so all of the flux passes through the side of the cube at x = a. This means that θ = 0º. Thus,

Φ = EA

Then, E = bx² and we're interested in the point where x = a, so if we substitute x for a, we have

E = ba²

Since A = a² for the cube face, we have

Q / ε₀ = ba² * a²

so

Q = ba⁴ * ε₀

You might be interested in

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the

Doss [256]

Answer:

$v_{su} = 19.44 m/s$

Explanation:

$m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg$

$T=9.29x10^8\\r_{o}=1.43x10^{12}$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}$

solve to r1

$r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = \frac{m_{sa}*2*pi*r_1^2}{T}$

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum. So that means

$L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}$

Since

$L_{su}= m_{su}*v_{su}*r_1$

then

$v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}$

$v_{su} = 19.44 m/s$

7 0

4 years ago

Do someone know the answer

JulijaS [17]

Speed x time = distance
Distance divided by time = speed
500 divided by 5
Speed = 100

4 0

3 years ago

A discus thrower turns with angular acceleration of 50 rad/s2, moving the discus in a circle of radius 0.80m. Find the radial an

anyanavicka [17]

Answer:

The value of tangential acceleration $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

The value of radial acceleration $\alpha_{r} = 80 \frac{m}{s^{2} }$

Explanation:

Angular acceleration = 50 $\frac{rad}{s^{2} }$

Radius of the disk = 0.8 m

Angular velocity = 10 $\frac{rad}{s}$

We know that tangential acceleration is given by the formula $\alpha_{t} =$ $r \alpha$

Where r = radius of the disk

$\alpha$ = angular acceleration

⇒ $\alpha_{t} =$ 0.8 × 50

⇒ $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

This is the value of tangential acceleration.

Radial acceleration is given by

$\alpha_{r} = \frac{V^{2} }{r}$

Where V = velocity of the disk = r $\omega$

⇒ V = 0.8 × 10

⇒ V = 8 $\frac{m}{s}$

Radial acceleration

$\alpha_{r} = \frac{8^{2} }{0.8}$

$\alpha_{r} = 80 \frac{m}{s^{2} }$

This is the value of radial acceleration.

7 0

3 years ago

represents the space-time, speed-time and acceleration-time graphs for a ball pulled upwards with a speed of 10 m / s from 1 met

alexandr1967 [171]

Gyhugugygubuhuhuhubu

5 0

3 years ago

A wave transfers:<br> Water<br> particles<br> energy<br> matter

SashulF [63]

Answer:

Particles in a water wave exchange kinetic energy for potential energy. When particles in water become part of a wave, they start to move up or down. This means that kinetic energy (energy of movement) has been transferred to them

Explanation:

hope this helps u ....

<em>pls mark this as the brainliest...</em>

6 0

3 years ago