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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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6 0
3 years ago
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Elodia [21]

Answer:

K_{i}+U_{g,i} = K_{f}+U_{g,f}

Explanation:

A closed system is a system where exists energy interactions with surroundings, but not mass interactions. If we neglect any energy interactions from boundary work, heat, electricity, magnetism and nuclear phenomena and assume that process occurs at steady state and all effects from non-conservative forces can be neglected, then the equation of energy conservation is reduce to this form:

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\Delta U_{g} - Change in gravitational potential energy of the system, measured in joules.

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K_{i}+U_{g,i} = K_{f}+U_{g,f} (2)

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7 0
3 years ago
Read 2 more answers
A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen
lara [203]

Answer:

frequency =  1475.45 Hz

Explanation:

given data

frequency f1 = 1215 Hz,

frequency f2 = 1265 Hz

police car moving vp = 25.0 m/s

solution

speed of sound u = 343 m / s

speed of the other car = v

when the police car is stationary

the frequency the other car receives is

f2 =  f1  ×  \dfrac{u+v}{u}      ................1

and

the frequency the police car receives is

 f2 =  f1  ×  \dfrac{u}{u-v}      ..................2

now from equation 1 and 2

\frac{f2}{f1} = \dfrac{u+v}{u-v}

 \frac{1275}{1240} = \frac{u+v}{u-v}

v =\frac{1275-1240}{1275+1240}\times 343  

v = 4.77 m/s

and

frequency the other car receives is  

f2 = f1 ×   \dfrac{u+v}{u-vp}       ......................3

and

the frequency the police car receives is

f2 = f1 ×  \dfrac{u+vp}{u - v}       .......................4

now we get

f2 = f1 ×  \dfrac{(u+v)(u + vp)}{(u-v)(u-vp)}      

f2 =    1240\times \frac{(343+4.77)(343+25)}{(343-4.77)(343-25)}        

f2 =  1475.45 Hz

 

4 0
3 years ago
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katovenus [111]

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The speed of the bullet is 217.43298 m/s

5 0
3 years ago
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