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iogann1982 [59]
3 years ago
13

Problem 1: A positively charged particle Q1 = +35 nC is held fixed at the origin. A second charge Q2 of mass m = 8.5 micrograms

is floating a distance d = 25 cm above charge Q1. The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth. A) What is the sign of charge on Q2? B) Calculate the magnitude of Q2 in units of nanocoulombs.
Physics
1 answer:
tigry1 [53]3 years ago
3 0

Answer:This problem has been solved! See the answer. Two charged particles, Q 1 = + 5.10 nC and Q 2 = -3.40 nC, are separated by 40.0 cm. (a) What is the ... has not been graded yet. (b) What is the electric potential at a point midway between the charged particles? (Note: Assume a reference level of potential V = 0 at r = infinity .)

Explanation:

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3 years ago
) The square plates of a 5000-pF parallel-plate capacitor measure 50 mm by 50 mm and are separated by a dielectric that is 0.23
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Answer:

4 x 10⁻⁴ J

Explanation:

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2 years ago
Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the
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Answer: 1.8\°

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by the following equation:

dsin\theta_{n}=n\lambda (1)

Where:

d=3.4(10)^{-5}m is the width of the slit

\lambda=540 nm=540(10)^{-9}m is the wavelength of the light  

n is an integer different from zero.

Now, the second-order diffraction angle is given when n=2, hence equation (1) becomes:

dsin\theta_{2}=2\lambda (2)

Now we have to find the value of \theta_{2}:

sin\theta_{2}=\frac{2\lambda}{d} (3)

Then:

\theta_{2}=arcsin(\frac{2\lambda}{d})   (4)

\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})   (5)

Finally:

\theta_{2}=1.8\°   (6)

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Explanation:

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Answer:

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Explanation:

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3 years ago
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