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sweet [91]
3 years ago
6

A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe

r face carries +95.0 nC/m2 while the lower face carries -25.0 nC/ m2. What is the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face? (ε0 = 8.85 × 10-12 C2/N · m2) A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The upper face carries +95.0 nC/m2 while the lower face carries -25.0 nC/ m2. What is the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face? ( = 8.85 × 10-12 C2/N · m2) 0.00 N/C 6.78 × 103 N/C 7.91 × 103 N/C 1.36 × 104 N/C 3.95 × 103 N/C
Physics
1 answer:
Nonamiya [84]3 years ago
4 0

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

=  surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ /  2ε₀

= 6.78 X 10³ N/C

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A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle
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Answer:

(a). Index of refraction are n_{red} = 1.344 & n_{violet} = 1.406

(b). The velocity of red light in the glass v_{red} = 2.23 ×10^{8} \ \frac{m}{s}

The velocity of violet light in the glass v_{violet} =2.13 ×10^{8} \ \frac{m}{s}

Explanation:

We know that

Law of reflection is

n_1 \sin\theta_{1} = n_2 \sin\theta_{2}

Here

\theta_1 = angle of incidence

\theta_2 = angle of refraction

(a). For red light

1 × \sin 56.6 = n_{red} × \sin 38.4

n_{red} = 1.344

For violet light

1 × \sin 56.6 = n_{violet} × \sin 36.4

n_{violet} = 1.406

(b). Index of refraction is given by

n = \frac{c}{v}

n_{red} = 1.344

v_{red} = \frac{c}{n_{red} }

v_{red} = \frac{3(10^{8} )}{1.344}

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This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

v_{violet} = \frac{3(10^{8} )}{1.406}

v_{violet} =2.13 ×10^{8} \ \frac{m}{s}

This is the velocity of violet light in the glass.

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Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

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(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

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Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

<u>k = 1684.38 N/m = 1.684 KN/m</u>

<u></u>

<u>(</u>c<u>)</u>

Applying Hooke's Law with:

Δx  = 0.6 m

F = (1684.38 N/m)(0.6 m)

<u>F = 1010.62 N = 1.01 KN</u>

<u></u>

(b)

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F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

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Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

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-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

<u>Vi = 0.105 m/s</u>

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