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LUCKY_DIMON [66]
3 years ago
10

You are working on a laboratory device that includes a small sphere with a large electric charge Q. Because of this charged sphe

re, there is a strong electric field surrounding your device. Other researchers in your laboratory are complaining that your electric field is affecting their equipment. You think about how you can obtain the large electric field that you need close to the sphere but prohibit the field from reaching your colleagues. You decide to surround your device with a spherical transparent plastic shell. The nonconducting shell is given a uniform charge distribution. Required:a. The shell is placed so that the small sphere is at the exact center of the shell. Determine the charge that must be placed on the shell to completely eliminate the electric field outside of the shell. b. What if the shell moves? Does the small sphere have to be at the center of the shell for this scheme to work?
Physics
1 answer:
madam [21]3 years ago
5 0

Answer:

the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.

Explanation:

We can answer this exercise using Gauss's law

      Ф = ∫ e . dA = q_{int} / ε₀

field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell.  the flow must be zero since the charge of the sphere is equal  induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field

From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.

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A proton moving along the positive x-axis enters a uniform magnetic field which is directed along the
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Answer: The force acting on the proton of charge ​1.5 x10-19 C moving with velocity 1.5 x10-19 C under the influence of a magnetic field of 0.7 T will be 3.15×10^-13 N.

Explanation: To find the answer we need to know more about the Lorentz magnetic force.

<h3>What is the Lorentz magnetic force acting on the proton?</h3>
  • Consider a proton of charge q moving with a velocity v in a magnetic field, then the Lorentz magnetic force exerted on the proton can be expressed as,

                                  F= q (v× B)

                               F= qvBsin\alpha  where, \alpha is the angle between v and B.

  • In the question, it is given that,

                                B=0.7 T\\q=1.5*10^{-19}C\\v= 3*10^{6}m/s.\\\alpha =90 degree.\\ because, from the question it is clear that the proton is moving along x axis and the magnetic field is along the y axis.

  • Thus, we can find the force acting on the proton as,

                      F=qvBsin\alpha =1.5*10^{-19}C*3*10^6 m/s*0.7T*sin (90)\\F=3.15*10^{-13}N

Thus, we can conclude that the Lorentz force acting on the proton will be 3.15×10^-13 N.

Learn more about the Lorentz magnetic force here:

brainly.com/question/28047923

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Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(
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y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude  = 2.20 mm

b) frequency =

     f = \dfrac{\omega}{2\pi}

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