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3 years ago

What happens to the speed of light as it passes into a glass block?

2 answers:

8 0

It slows down. The reason for this is because when the light travels through the air, it is at it’s maximum speed. When it passes through the glass block, it slows down because the medium is denser than air. This also causes refraction.

Deffense [45]3 years ago

7 0

<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ The speed would reduce as glass is denser than air.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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3 years ago

What is the magnitude of the electric field at a point midway between a −5.6μC and a +5.8μC charge 9.0cm apart? Assume no other

Rashid [163]

Answer:

Magnitude of electric field at mid way is $5.07\times 10^{7} N/C$

Solution:

As per the question:

Q = $- 5.6\mu m = - 5.6\times 10^{- 6} C$

Q' = $5.8\mu m = 5.8\times 10^{- 6} C$

Separation distance, d = 9.0 cm = 0.09 m

The distance between charges at mid-way, O is $\farc{d}{2} = 0.045 m$

Now, the electric at point O due to charge Q is:

$E = \frac{1}{4\pi \epsilon_{o}}.\frac{Q}{(\frac{d}{2})^{2}}$

$E = \frac{1}{4\pi \epsilon_{o}}.\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}$

$E = (9\times 10^{9})\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}$

$E = - 2.49\times 10^{7} N/C$

Here, negative sign is indicative of the direction of electric field which is towards the point O

Now, the electric at point O due to charge Q' is:

$E' = \frac{1}{4\pi \epsilon_{o}}.\frac{Q}{(\frac{d}{2})^{2}}$

$E' = (9\times 10^{9})\frac{5.8\times 10^{- 6}}{(0.045^{2}}$

$E' = 2.58\times 10^{7} N/C$

Refer to Fig 1.

Since, both the fields are in the same direction:

$E_{net} = E + E' = 2.49\times 10^{7} + 2.58\times 10^{7} = 5.07\times 10^{7} N/C$

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Art [367]

Answer:

I think the answer is B.

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hop it helps

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Ostrovityanka [42]

D I took the quiz and I'm pretty sure you will get it correct

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