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4vir4ik [10]
3 years ago
11

There is an electric field in the region between the two plates. The magnitude of this electric field is ed. This imposes anothe

r condition on the charge densities on the surfaces of the plates. How can this condition be expressed?
Physics
1 answer:
krok68 [10]3 years ago
5 0

Answer:

it is essential that the charge on the plates are of the same magnitude, but in the opposite direction

Explanation:

The configuration of parallel plates is called a capacitor and is widely used to create constant electric fields inside.

 To obtain this field it is essential that the charge on the plates are of the same magnitude, but in the opposite direction

This is so that the fields created by each plate can be added inside and subtracted from the outside of the plates

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a current will be induced.

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What type of wave did you create when you pushed your arm back and forth?<br><br> plz help
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Are sound waves transverse waves or longitudinal waves? Why?
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We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes
Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

1.56421\times 10^{-6}\ T

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = \frac{d}{2}=\frac{7}{2}=3.5\ cm

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

c = Speed of light = 3\times 10^8\ m/s

The intensity is given by

I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2

5% of this energy goes to the visible light so the intensity is

I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

u=\frac{1}{2}\epsilon_0E^2

Energy density is also given by

\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T

The amplitude of the magnetic field at this surface is 1.56421\times 10^{-6}\ T

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3 years ago
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