Physics, 14.09.2019 08:30 10040816

An electric field of intensity 3.22 kn/c is applied along the x-axis. calculate the electric flux through a rectangular plane 0.350 m wide and 0.700 m long if the following conditions are true. (a) the plane is parallel to the yz. plane. (b) the plane is parallel to the xy. plane. (c) the plane contains the yaxis, and its normal makes an angle of 34.2* with the x-axis. the electric field everywhere on the surface of a charged sphere of radius 0.205 m has a magnitude of 570 n/c and points radially outward from the center of the sphere. (a) what is the net charge on the sphere? nc (b) what can you conclude about the nature and distribution of charge inside the sphere?

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Answer from: skykopystecky8769

Part I:

(a): $7.889\times 10^2\ Nm.$

(b): $0\ Nm.$

(c): $6.52\times 10^2\ Nm.$

Part II:

(a): $2.664\times 10^{-9}\ C.$

(b): Charge on the sphere is positive and only distributed on the surface of the sphere and not inside the sphere.

Explanation:

Part I:

Assuming,

$\hat i,\ \hat j,\ \hat k$ are the unit vectors along positive x, y and z axes respectively.

Given, the electric field is of intensity 3.22 kN/C and is along x-axis.

Therefore,

$\vec E = 3.22\ kN/C\ \hat i=3.22\times 10^3\ N/C\ \hat i.$

The magnetic flux through a surface is defined as

$\phi = \vec E\cdot \vec A$

where,

$\vec A$ is the area vector of the surface which is directed along the normal to the plane of the surface and its magnitude is equal to the area of the surface.

$A = \text{length of the plane }\times \text{width of the plane}\\=0.700\times 0.350\\=0.245\ m^2.$

(a):

When the plane is parallel to yz plane, then its normal is along x axis, therefore,

$\vec A = A\ \hat i.$

Electric flux,

$\phi = \vec E\cdot \vec A \\=(3.22\times 10^3)\hat i\cdot (0.245)\hat i = 3.22\times10^3\times 0.245\times (\hat i\cdot \hat i) = 7.889\times 10^2\ Nm.$

(b):

When the plane is parallel to xy plane, then its normal is along z axis, therefore,

$\vec A = A\ \hat k.$

Electric flux,

$\phi = \vec E\cdot \vec A \\=(3.22\times 10^3)\hat i\cdot (0.245)\hat k = 3.22\times10^3\times 0.245\times (\hat i\cdot \hat k) = 0\ Nm.$

(c):

When the normal to the plane makes an angle $34.2^\circ$ with the x-axis.

Electric flux through the plane is given by,

$\phi = \vec E\cdot \vec A\\=EA\cos(34.2^\circ) \\=(3.22\times 10^3)\times (0.245)\times \cos(34.2^\circ)\\=6.52\times 10^2\ Nm.$

Part II:

(a):

Given:

Electric field everywhere on the surface of the sphere, $E = 570\ N/C.$ Radius of the sphere, $R = 0.205\ m.$

The electric field is directed radially outward from the surface of the sphere at its every point and the normal to the surface of the sphere is also radially outwards at its every point therefore both the electric field and the area vector of the sphere is along the same direction.

The flux through the surface is given by

$\phi = \vec E \cdot \vec A \\= EA\cos0^\circ\\=EA\\=E\ 4\pi R^2\\=570\times 4\pi \times 0.205^2\\=3.01\times 10^2\ Nm.$

According to Gauss law of electrostatics,

$\phi = \dfrac q{\epsilon_o}$

where,

= charge enclosed by the sphere. $\epsilon_o$ = electric permittivity of free space = $8.85\times10^{-12}\ C^2N^{-1}m^{-2}.$

Therefore, the net charge on the sphere is given by

$q=\epsilon_o\times \phi\\=8.85\times 10^{-12}\times 3.01\times 10^2\\=2.664\times 10^{-9}\ C.$

(b):

Since, the sphere is charged all of the charge resides on its surface. The electric field through the sphere is radially outwards which means the charge on the sphere should be positive.

There is no charge distributed inside the sphere.

Answer from: Quest

the answer is kevlar

Answer from: Quest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

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An electric field of intensity 3.22 kn/c is applied along the x-axis. calculate the electric flux th...

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