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leva [86]
3 years ago
9

If the electric field just outside a thin conducting sheet is equal to 1.5 N/C, determine the surface charge density on the cond

uctor.
Engineering
1 answer:
Dennis_Churaev [7]3 years ago
7 0

Answer:

The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²

Explanation:

The electric field intensity due to a thin conducting sheet is given by the following formula:

Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)

From this formula:

Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)

We have the following data:

Electric Field Intensity = 1.5 N/C

Permittivity of free space = 8.85 x 10^-12 C²/N.m²

Therefore,

Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)

<u>Surface Charge Density = 26.55 x 10^-12 C/m²</u>

Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².

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Answer:

Explanation:

From the equation:

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A thermoelectric refrigerator is powered by a 16-V power supply that draws 2.9 A of current. If the refrigerator cools down 3.1
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Answer:

COP = 0.090

Explanation:

The general formula for COP is:

COP = Desired Output/Required Input

Here,

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Desired Output = (Specific Heat of Water)(Mass of Water)(Change in Temperature)/Time

Desired Output = [(4180 J/kg.k)(3.1 kg)(25 - 11)k]/[(12 hr)(3600 sec/hr)]

Desired Output = 4.199 W

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A DC generator turns at 2000 rpm and has an output of 200 V. The armature constant is 0.5 V-min/Wb, and the field constant of th
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