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2 years ago

9

Where is visible light located on the electromagnetic spectrum?

2 answers:

Stells [14]2 years ago

7 0

Answer:

2. Near the middle

Explanation:

Visible light is a form of electromagnetic radiation which is visible to the human eye, it is located in the narrow band of wavelengths found on the left of the spectrum, with the ultraviolet wavelengths to the left and infrared to the right side of the spectrum. This places the visible light near the middle of the spectrum.

The human eyes have high sensitivity for detecting very narrow band of wavelengths despite the spectrum having a wide array of wavelengths. Visible light ranges from 400 nm to 700 nm

nikklg [1K]2 years ago

4 0

2 near the middle
between ultraviolet and infared

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Find electric field at point p which is a distance l away from the both +q and -q

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Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

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