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Nata [24]
3 years ago
5

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?
Physics
1 answer:
Arte-miy333 [17]3 years ago
7 0

Answer: 7.91 * 10^3 N/C

Explanation: In order to solve this problem we have to use the Gaussian law, in both charged surface for the infinite plane.

So inside the non conducting sheet we apply teh superposition principle adding the electric field from each charged surface.

Then we have

Eupper=σ+/εo

Ebottom= σ-/εo

Adding the electric fields we have:

E inside= Eupper-Ebottom=(1/εo)*(σ+-σ-)= (1/8.85* 10^-12)*70 nC/m^2= 7.91* 10^3N/C

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BARSIC [14]

Answer:

<em>The non resonance frequency of the generator is = 1201.79 Hz</em>

Explanation:

At resonance,

f₀ = 1/2π√LC..................... Equation 1

Where f₀ = resonance frequency, L = inductance, C = capacitance

making LC the subject of the equation

LC = 1/4πf₀²..................... Equation 2

<em>Given: </em>f₀ = 225 Hz, and π = 3.143

<em>Substituting these values into equation 2,</em>

LC = 1/(4×3.143²×225²)

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If the ratio of capacitive reactance to inductive reactance = 5.36

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Where f = frequency of the non resonant

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Substituting the value of LC = 5×10⁻⁷ into equation 3

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f = 5.36/(6.286×0.00071)

f = 5.36/0.00446

<em>f = 1201.79 Hz</em>

<em>Thus the non resonant frequency of the generator is = 1201.79 Hz</em>

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