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Blababa [14]
4 years ago
6

A +17 nc charge is located at the origin. what is the strength of the electric field at the position (x,y)=(5.0cm,0cm)? express

your answer using
Physics
1 answer:
gavmur [86]4 years ago
7 0
The working equation to be used for this is written below:

E = kQ/d²
where
E is the electric field
k is a constant equal to 8.99 x 10⁹ N m²/C²
Q is the charge
d is the distance


E = (8.99 x 10⁹ N m²/C²)(17×10⁻⁹ C)/(0.05 m)²
E = 61,132 N/C
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Answer:

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Explanation:

We can use the Young's Double Slit Experiment Formula here:

\Delta x = \frac{\lambda L}{d}\\\\

where,

Δx = distance between consecutive dark fringes = width of central bright fringe = ?

λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m

L = distance between screen and slit = 3.7 m

d = slit width = 0.37 mm = 3.7 x 10⁻⁴ m

Therefore,

\Delta x = \frac{(6.33\ x\ 10^{-7}\ m)(3.7\ m)}{3.7\ x \ 10^{-4}\ m}

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Answer:

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