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Lyrx [107]
3 years ago
10

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

.7 μC/m2. A thick, infinite conducting slab, also oriented perpendicular to the x-axis occupiees the region between a = 2.8 cm and b = 4.8 cm. The conducting slab has a net charge per unit area of σ2 = 88 μC/m2. What is Ex(P), the value of the x-component of the electric field at point P, located a distance 6.8 cm from the infinite sheet of charge?
Physics
2 answers:
pishuonlain [190]3 years ago
7 0

Answer:

E_{total}=4.82*10^6N/C

vector with direction equal to the axis X.

Explanation:

We use the Gauss Law and the superposition law in order to solve this problem.

<u>Superposition Law:</u> the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

E_{total}=E_1+E_2

<u>Thanks Gauss Law</u> we know that the electric field of a infinite sheet with density of charge σ is:

E=\sigma/(2\epsilon_o)

Then:

E_{total}=(\sigma_1+\sigma_2)/(2\epsilon_o)=(-2.7*10^{-6}+88*10^{-6})/(2*8.85*10^{-12})=4.82*10^6N/C

This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.

slava [35]3 years ago
6 0

Answer:

The total electric field will be 5*10^5 N/C

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Given:

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Now,

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Now, using eqn (1), we get:

\frac{v_{b}}{v_{a}} = \frac{\rho_{a}}{\rho_{b}}

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3 years ago
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In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10
sergij07 [2.7K]

Answer:

A)\Phi=83.84\times 10^{-9}

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Explanation:

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(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

\Phi=B.a\,cos \theta..................................(1)

\theta is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

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(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

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\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}

\Phi=0 Wb

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According to the Faraday's Law we have:

emf=n\frac{B.a}{t}

emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}

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Answer:

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In this problem, we notice that:

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