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3 years ago

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An electron in a vacuum chamber is fired with a speed of 7400 km/s toward a large, uniformly charged plate 75 cm away. The elect

ron reaches a closest distance of 15 cm before being repelled.
What is the plate's surface charge density?

1 answer:

Klio2033 [76]3 years ago

4 0

Answer:

2.29e-9C/m²

Explanation:

Using E = σ/ε₀ means the force on the electron is F = eE = eσ/ε₀.

The work done on the electron is W = Fd = deσ/ε₀. This equals the kinetic energy lost, ½mv².

½mv² = deσ/ε₀

d = 75cm – 15cm = 60cm = 0.6m

σ = mv²ε₀/(2de)

. .= 9.11e-31 * (7.4e6)² * 8.85e-12 / (2 * 0.6 * 1.6e-19)

. .= 2.29e-9 C/m² (i.e. 2.29x10^-9 C/m²)

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Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2

aleksandr82 [10.1K]

Answer:

The curl is $0 \hat x -z^2 \hat y -4xy \hat z$

Explanation:

Given the vector function

$\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z$

We can calculate the curl using the definition

$\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|$

Thus for the exercise we will have

$\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|$

So we will get

$\nabla \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z$

Working with the partial derivatives we get the curl

$\nabla \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z$

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3 years ago

A -turn rectangular coil with length and width is in a region with its axis initially aligned to a horizontally directed uniform

madam [21]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The maximum emf is $\epsilon_{max}= 26.8 V$

The emf induced at t = 1.00 s is $\epsilon = 24.1V$

The maximum rate of change of magnetic flux is $\frac{d \o}{dt}|_{max} =26.8V$

Explanation:

From the question we are told that

The number of turns is N = 44 turns

The length of the coil is $l = 15.0 cm = \frac{15}{100} = 0.15m$

The width of the coil is $w = 8.50 cm =\frac{8.50}{100} =0.085 m$

The magnetic field is $B = 745 \ mT$

The angular speed is $w = 64.0 rad/s$

Generally the induced emf is mathematically represented as

$\epsilon = \epsilon_{max} sin (wt)$

Where $\epsilon_{max}$ is the maximum induced emf and this is mathematically represented as

$\epsilon_{max} = N\ B\ A\ w$

Where $\o$ is the magnetic flux

N is the number of turns

A is the area of the coil which is mathematically evaluated as

$A = l *w$

Substituting values

$A = 0.15 * 0.085$

$= 0.01275m^2$

substituting values into the equation for maximum induced emf

$\epsilon_{max} = 44* 745 *10^{-3} * 0.01275 * 64.0$

$\epsilon_{max}= 26.8 V$

given that the time t = 1.0sec

substituting values into the equation for induced emf $\epsilon = \epsilon_{max} sin (wt)$

$\epsilon = 26.8 sin (64 * 1)$

$\epsilon = 24.1V$

The maximum induced emf can also be represented mathematically as

$\epsilon_{max} = \frac{d \o}{dt}|_{max}$

Where $\o$ is the magnetic flux and $\frac{d \o}{dt}|_{max}$ is the maximum rate at which magnetic flux changes the value of the maximum rate of change of magnetic flux is

$\frac{d \o}{dt}|_{max} =26.8V$

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