The formula that will be used in this problem is E = q/ 4pi*r^2 z where z is the elctric charge constant equal to 8.854 *10 ^-12. The magnitude using r equal to 0.0525 m and q equal to -22.3 *10^-6 C is equal to -22.3 *10^-6/ 4pi*(0.0525)^2 *8.854 *10 ^-12 or equal to -7.272 *10 ^7. The magnitude 5 cm outside the surface is -22.3 *10^-6<span>/ 4pi*(0.0525+0.05)^2 *8.854 *10 ^-12 equal to -1.908 *10^7.
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For displacement apply Pythagorean theorem
Answer:
The correct answer is Gamma Rays.
Explanation:
Answer:
0.532
Explanation:
Your equation to find the second bright interference maximum is gonna be this: d sin (Θ) = m λ
First, find your variables.
λ = 580 · 10^-9
d = 0.000125
m = 2
Next, fill in the equation.
d sin (θ) = m λ
(0.000125) sin (θ) = (2) (580·10^-9)
Then isolate your variable.
θ = arcsin ( (2)(580·10^-9) / (0.000125) )
Run your equation and you will end up with 0.53171246 , which rounds to 0.532.
The main thing you have to watch out for is make sure you are calculating for the bright interference and not the dark interference, as well as checking you're calculating for the maximum, not the minimum.
I hope this helps :D